max min word problems - department of mathematicsbill/meannapps/l5maxminwordproblemsbeamer.pdf6....
TRANSCRIPT
![Page 1: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/1.jpg)
Max Min Word Problems
Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
3. write down the information of the problem in terms of thoseletters;
4. write down other relevant facts;
5. restate and solve the problem;
6. guarantee that our solution is as claimed.
![Page 2: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/2.jpg)
Max Min Word Problems
Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
3. write down the information of the problem in terms of thoseletters;
4. write down other relevant facts;
5. restate and solve the problem;
6. guarantee that our solution is as claimed.
![Page 3: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/3.jpg)
Max Min Word Problems
Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
3. write down the information of the problem in terms of thoseletters;
4. write down other relevant facts;
5. restate and solve the problem;
6. guarantee that our solution is as claimed.
![Page 4: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/4.jpg)
Max Min Word Problems
Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
3. write down the information of the problem in terms of thoseletters;
4. write down other relevant facts;
5. restate and solve the problem;
6. guarantee that our solution is as claimed.
![Page 5: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/5.jpg)
Max Min Word Problems
Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
3. write down the information of the problem in terms of thoseletters;
4. write down other relevant facts;
5. restate and solve the problem;
6. guarantee that our solution is as claimed.
![Page 6: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/6.jpg)
Max Min Word Problems
Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
3. write down the information of the problem in terms of thoseletters;
4. write down other relevant facts;
5. restate and solve the problem;
6. guarantee that our solution is as claimed.
![Page 7: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/7.jpg)
Max Min Word Problems
Our approach to max min word problems is modeled after ourapproach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
3. write down the information of the problem in terms of thoseletters;
4. write down other relevant facts;
5. restate and solve the problem;
6. guarantee that our solution is as claimed.
![Page 8: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/8.jpg)
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?
y
x
x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.
We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.
x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.
![Page 9: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/9.jpg)
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?
y
x
x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.
We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.
x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.
![Page 10: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/10.jpg)
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?
y
x
x + 2y = 100
Area = A = xy =(100− 2y)y =100y − 2y2.
We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.
x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.
![Page 11: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/11.jpg)
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?
y
x
x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.
We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.
x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.
![Page 12: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/12.jpg)
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?
y
x
x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.
We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.
x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.
![Page 13: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/13.jpg)
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?
y
x
x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.
We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.
x ≥ 0 and y ≥ 0.
Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.
![Page 14: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/14.jpg)
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?
y
x
x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.
We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.
x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.
So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.
![Page 15: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/15.jpg)
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?
y
x
x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.
We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.
x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y .
That is, 0 ≤ y ≤ 50.
![Page 16: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/16.jpg)
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangulargarden. One side will be against a long barn. He has 100 ft offencing that he will use to surround the other three sides. Whatare the dimensions of the garden of maximum area?
y
x
x + 2y = 100Area = A = xy =(100− 2y)y =100y − 2y2.
We wish tofind the maximum valueof A and the values of xand y that produce thatmaximum value of A.
x ≥ 0 and y ≥ 0. Since x = 100− 2y , we see that 100− 2y ≥ 0.So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.
![Page 17: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/17.jpg)
Biggest Garden
So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.
These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA
dy= 100− 4y .
A is differentiable everywhere, anddA
dy= 0 iff y = 25.
If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.
![Page 18: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/18.jpg)
Biggest Garden
So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.
dA
dy= 100− 4y .
A is differentiable everywhere, anddA
dy= 0 iff y = 25.
If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.
![Page 19: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/19.jpg)
Biggest Garden
So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA
dy= 100− 4y .
A is differentiable everywhere, anddA
dy= 0 iff y = 25.
If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.
![Page 20: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/20.jpg)
Biggest Garden
So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA
dy= 100− 4y .
A is differentiable everywhere, anddA
dy= 0 iff y = 25.
If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.
![Page 21: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/21.jpg)
Biggest Garden
So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA
dy= 100− 4y .
A is differentiable everywhere, anddA
dy= 0 iff y = 25.
If y = 0, then A = 0.
If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.
![Page 22: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/22.jpg)
Biggest Garden
So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA
dy= 100− 4y .
A is differentiable everywhere, anddA
dy= 0 iff y = 25.
If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.
If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.
![Page 23: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/23.jpg)
Biggest Garden
So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA
dy= 100− 4y .
A is differentiable everywhere, anddA
dy= 0 iff y = 25.
If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.
So the maximum area occurs with y = 25 and x = 100− 2y = 50.
![Page 24: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/24.jpg)
Biggest Garden
So we use our techniques for finding the maximum value of afunction on a closed interval when the function is continuous there.These apply since 100y − 2y2 is a polynomial and so continuouseverywhere.dA
dy= 100− 4y .
A is differentiable everywhere, anddA
dy= 0 iff y = 25.
If y = 0, then A = 0.If y = 25, then A = 100 · 25− 2 · 252 = (100− 50)25 > 0.If y = 50, then x = 100− 2 · 50 = 0, and A = 0 · 50 = 0.So the maximum area occurs with y = 25 and x = 100− 2y = 50.
![Page 25: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/25.jpg)
Closest Point
Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.
d
xK3 K2 K1 0 1 2 3
y
K1
1
2
3
4
5
6
We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =
√(x − 3)2 + (y − 0)2 =
√(x − 3)2 + y2.
We want to find the point 〈x , y〉that gives the minimum value for the distance d .
RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.
![Page 26: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/26.jpg)
Closest Point
Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.
d
xK3 K2 K1 0 1 2 3
y
K1
1
2
3
4
5
6
We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉.
Sod =
√(x − 3)2 + (y − 0)2 =
√(x − 3)2 + y2.
We want to find the point 〈x , y〉that gives the minimum value for the distance d .
RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.
![Page 27: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/27.jpg)
Closest Point
Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.
d
xK3 K2 K1 0 1 2 3
y
K1
1
2
3
4
5
6
We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =
√(x − 3)2 + (y − 0)2 =
√(x − 3)2 + y2.
We want to find the point 〈x , y〉that gives the minimum value for the distance d .
RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.
![Page 28: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/28.jpg)
Closest Point
Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.
d
xK3 K2 K1 0 1 2 3
y
K1
1
2
3
4
5
6
We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =
√(x − 3)2 + (y − 0)2 =
√(x − 3)2 + y2.
We want to find the point 〈x , y〉that gives the minimum value for the distance d .
RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.
![Page 29: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/29.jpg)
Closest Point
Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.
d
xK3 K2 K1 0 1 2 3
y
K1
1
2
3
4
5
6
We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =
√(x − 3)2 + (y − 0)2 =
√(x − 3)2 + y2.
We want to find the point 〈x , y〉that gives the minimum value for the distance d .
RemarkWe let s = d2. So s = (x − 3)2 + y2.
Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.
![Page 30: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/30.jpg)
Closest Point
Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.
d
xK3 K2 K1 0 1 2 3
y
K1
1
2
3
4
5
6
We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =
√(x − 3)2 + (y − 0)2 =
√(x − 3)2 + y2.
We want to find the point 〈x , y〉that gives the minimum value for the distance d .
RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?)
So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.
![Page 31: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/31.jpg)
Closest Point
Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.
d
xK3 K2 K1 0 1 2 3
y
K1
1
2
3
4
5
6
We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =
√(x − 3)2 + (y − 0)2 =
√(x − 3)2 + y2.
We want to find the point 〈x , y〉that gives the minimum value for the distance d .
RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d .
This approach isfrequently useful in max min distance problems.
![Page 32: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/32.jpg)
Closest Point
Example 2. Problem. Find the point on y = x2 closest to thepoint 〈3, 0〉.
d
xK3 K2 K1 0 1 2 3
y
K1
1
2
3
4
5
6
We let 〈x , y〉 be a point on the curve, and we letd be the distance between 〈x , y〉 and 〈3, 0〉. Sod =
√(x − 3)2 + (y − 0)2 =
√(x − 3)2 + y2.
We want to find the point 〈x , y〉that gives the minimum value for the distance d .
RemarkWe let s = d2. So s = (x − 3)2 + y2. Theminimum value for s and the minimum value ford will occur at the same point. (Why?) So wewill work with s instead of d . This approach isfrequently useful in max min distance problems.
![Page 33: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/33.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.
Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 34: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/34.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 35: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/35.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 36: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/36.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 =
2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 37: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/37.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) =
4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 38: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/38.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 39: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/39.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) =
4(x − 1)((x + 12)2 + 5
4).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 40: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/40.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 41: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/41.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3.
Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 42: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/42.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0.
So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 43: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/43.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3.
We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 44: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/44.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division.
Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 45: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/45.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 46: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/46.jpg)
Closest Point
Now s = (x − 3)2 + y2 = (x − 3)2 + x4 since y = x2.Note that s is a polynomial and so continuous everywhere.As usual, we compute the derivative of s.
Sods
dx= 2(x − 3) + 4x3 = 2(2x3 + x − 3) =
2(x − 1)(2x2 + 2x + 3) = 4(x − 1)(x2 + x + 32) =
4(x − 1)(x2 + 2(12)x + 1
4 + 54) = 4(x − 1)((x + 1
2)2 + 54).
RemarkLet g(x) = 2x3 + x − 3. Note that g(1) = 0. So (x − 1) divides2x3 + x − 3. We find the quotient by long division. Also, we findthat x2 + x + 3
2 = (x + 12)2 + 5
4 by “completing the square”.
![Page 47: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/47.jpg)
Closest Point
(x + 12)2 + 5
4 ≥54 > 0 for all x .
So if x < 1, then x − 1 < 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) < 0.
So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) > 0.
So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.
![Page 48: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/48.jpg)
Closest Point
(x + 12)2 + 5
4 ≥54 > 0 for all x .
So if x < 1, then x − 1 < 0,
and sods
dx= 4(x − 1)((x + 1
2)2 + 54) < 0.
So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) > 0.
So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.
![Page 49: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/49.jpg)
Closest Point
(x + 12)2 + 5
4 ≥54 > 0 for all x .
So if x < 1, then x − 1 < 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) < 0.
So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) > 0.
So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.
![Page 50: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/50.jpg)
Closest Point
(x + 12)2 + 5
4 ≥54 > 0 for all x .
So if x < 1, then x − 1 < 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) < 0.
So s is decreasing on (−∞, 1].
So if x > 1, then x − 1 > 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) > 0.
So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.
![Page 51: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/51.jpg)
Closest Point
(x + 12)2 + 5
4 ≥54 > 0 for all x .
So if x < 1, then x − 1 < 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) < 0.
So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0,
and sods
dx= 4(x − 1)((x + 1
2)2 + 54) > 0.
So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.
![Page 52: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/52.jpg)
Closest Point
(x + 12)2 + 5
4 ≥54 > 0 for all x .
So if x < 1, then x − 1 < 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) < 0.
So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) > 0.
So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.
![Page 53: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/53.jpg)
Closest Point
(x + 12)2 + 5
4 ≥54 > 0 for all x .
So if x < 1, then x − 1 < 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) < 0.
So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) > 0.
So s is increasing on [1,∞).
So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.
![Page 54: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/54.jpg)
Closest Point
(x + 12)2 + 5
4 ≥54 > 0 for all x .
So if x < 1, then x − 1 < 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) < 0.
So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) > 0.
So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.
So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.
![Page 55: Max Min Word Problems - Department of Mathematicsbill/MEANnAPPS/L5MaxMinWordProblemsBeamer.pdf6. guarantee that our solution is as claimed. ... Our approach to max min word problems](https://reader031.vdocuments.net/reader031/viewer/2022030420/5aa6c8757f8b9a50528b74de/html5/thumbnails/55.jpg)
Closest Point
(x + 12)2 + 5
4 ≥54 > 0 for all x .
So if x < 1, then x − 1 < 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) < 0.
So s is decreasing on (−∞, 1].So if x > 1, then x − 1 > 0, and sods
dx= 4(x − 1)((x + 1
2)2 + 54) > 0.
So s is increasing on [1,∞).So the absolute minimum value for s (and d) occurs with x = 1.So 〈1, 1〉 is the point on y = x2 closest to 〈3, 0〉.