4.7 FORMING FUNCTIONS FROM VERBAL DESCRIPTIONS

Objectives: 1) Form a function in one variable from a verbal description

2) Use the graphing calculator to approximate a value for a graphic solution to a verbal problem

Question 1A pile of sand is in the shape of a cone with a diameter that is twice its height. Sketch the diagram. Express the volume V as a function of the height of the pile. Since the volume of a cone is given by

V = 1/3r2h

and d = 2h, therefore 2r = 2h or h = r, we have

V = 1/3h3

In this last equation, the volume is a function of the height of the sand pile (cone).

Strategy

1) Identify given information.

2) Assign variables to the related quantities.

3) Use formula(s) to link all variables together (Primary Equation).

4) Find out the relationship among other variables to one variable (Secondary Equation(s)).

5) Rewrite the formula(s) again in terms of one variable.

Question 2 A winch at the tip of a 12-meter building pulls pipe of the same length to a vertical position. The length of a rope at top of the building holds the pipe. Find the length of the rope in terms of the height.

s(x, y)

12

12

[Solution] The relationships among the three variables are:

222 12 yx 2 2 2(12 )s x y and

2 2144x y 2 2 2144 (12 )s y y

2 2 2144 144 24 288 24s y y y y

288 24s y

Question 3 An open top box with a square base is to be constructed from sheet metal in such a way that the completed box is made from 2 square yards of sheet metal.

1) Express the volume of the box as a multivariable function of the base and width of the square base dimensions.

2) Express the volume of the box as a single function of the width of the square base.

3) Find the feasible domain of the side of the base.

hxhxV 2),(

[Solution] (1)Let the length of the base be x inch,

xx

hand the height be h inch.

Then the volume of the box in terms of x and h inch is

Primary Equation

(2) Notice that “made from a 2 yd2 sheet metal” provides2( , ) 4 2S x h x xh Secondary

Equation

2(2 ) / 4h x x From the secondary equation, we have

Then the volume of the box in terms of x in the primary equation is

2 2 2( ) (2 ) / 4 (2 ) / 4V x x x x x x

(3) Since the height h > 0, then we need to solve the rational inequality

220

4

x

x

This results in

24 (2 ) 0

0

x x

x

0 2x

Question 4 A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume?

hxhxV 2),(

Solution

Let the length of the base be x inch,

xx

hand the height be h inch.

Then the volume of the box in terms of x and h inch is

Primary EquationNotice that “surface area of 108

in2 ” provides 1084),( 2 xhxhxS Secondary Equation

xxh 4/)108( 2

Solution

From the secondary equation, we have

xx

h

Then the volume of the box in terms of x in the primary equation is

Now, before we aim on V(x) and try to find its maximum volume while x has certain length, we should find the feasible domain of x.

4/274/)108()( 322 xxxxxxV

Question 4 A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume?

Notice that 22 4108),( xxhxhxS

This means 1080 x

0xand

Feasible Domain

Therefore, V(6, 3) is maximum when x = 6 and h = 3. The box dimension is 6 x 6 x 3.

Question 5 Which points on the graph of y = 4 – x2 are closest to the point (0, 2)?

x

y

d (x, y)

Solution

The distance between the point (0, 2) to any point (x, y) on the graph is 2222 )2()2()0( yxyxd

Notice that y = 4 – x2, then

43)24()( 24222 xxxxxd

Also notice that the x value makes the d(x) reaches the minimum is the same as makes the d2(x) reaches the minimum.

Question 5 Which points on the graph of y = 4 – x2 are closest to the point (0, 2)?

Solution

Therefore,

When

22 4 2 2 3 7( ) 3 4

2 4d x x x x

The d2(x) reaches the minimum. The minimum value is

3 6

2 2x

2min

7( )

4d x

So the points are closest to (0, 2) and the distance is

min

7( )

2d x

2 3

2x or and

3 54

2 2y

6 5,

2 2

Question 6 From a raft 50 m offshore, a lifeguard wants to swim to shore and run to a snack bar 100 m down the beach, as show.

a)If the lifeguard swims at 1 m/s and run at 3 m/s, express the total swimming and running time t as a function of the distance x show in the diagram?

b)Find the minimum time.

x100

Solution

50

The swimming distance in terms of x is

2 2( ) 50s x x The running distance in terms of x is ( ) 100r x x The total time in second in terms of x is

2 2 100( ) 50

3

xT x x

Question 7 A man is in boat 2 miles from the nearest point R on the coast. His is to go to a point Q, located 3 miles down the coast and 1 mile in land. He can row at 2 miles per hour and walk at 4 miles per hour. Toward what point on the coast should he row in order to reach point Q in the least time?

Q

2

1

3

Solution

Let the point on the coast toward by the man be P, and PR = x, then PS = 3 – x. Then

PR Sx

3 – x

M

42 xMP

1)3( 2 xPQ

Q

2

1

3

Then the total time from point M to Q will be

PR Sx

3 – x

M

4

106

2

4)(

22

xxxxT

The feasible domain for x is [0, 3]. The solution for x within the domain is x = 1.

Question 8 Four feet wire is to be used to form a square and a circle. How much of the wire should be used for the square and how much should be used for the circle to enclose the maximum area?Solution

r

x

x

Let the side of the square be x and the radius of the circle be r. Then

Since the total length of wire is 4 feet, then

22),( rxrxA Primary Equation

Secondary Equation

424 rx So,

)1(2 xr

r

x

x

The feasible domain of x is restricted by the square’s perimeter

22 )1(2

)(

xxxA

)1(2 x

r

22 )1(4 x

x

]48)4[(1 2 xx

10 x

We conclude that when yields the maximum area. That means all the wire is used to form circle.

0x

Question 9 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola

y = 4 – x2 that is above the x-axis.

a)Express the perimeter P of the rectangle as a function of the x-coordinate of A.

b)What is the domain of the perimeterfunction?

c) For what value of x is the perimetera maximum?

x

y

B A (x, y)

C D

Solution (a) ( , ) 2(2 ) 2 4 2P x y x y x y

2

2

( ) 4 2(4 )

2 4 8

P x x x

x x

Question 9 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola

y = 4 – x2 that is above the x-axis.

a)Express the perimeter P of the rectangle as a function of the x-coordinate of A.

b)What is the domain of the perimeterfunction?

c) For what value of x is the perimetera maximum?

x

y

B A (x, y)

C D

Solution (b)

Domain of the perimeter is0 2x

Question 9 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola

y = 4 – x2 that is above the x-axis.

a)Express the perimeter P of the rectangle as a function of the x-coordinate of A.

b)What is the domain of the perimeterfunction?

c) For what value of x is the perimetera maximum?

x

y

B A (x, y)

C D

Solution (c)

From the expression in (a), 2( ) 2 4 8P x x x

when x = 1, P(x) has a maximum.

max (1) 10P

Question 10 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola

y = 4 – x2 that is above the x-axis.

a)Express the area A of the rectangle as a function of the x-coordinate of A.

b)What is the domain of the areafunction?

c) For what value of x is the areaa maximum?Solution

(a) ( , ) 2A x y xy2( ) 2 (4 )A x x x

x

y

B A (x, y)

C D

Question 10 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola

y = 4 – x2 that is above the x-axis.

a)Express the area A of the rectangle as a function of the x-coordinate of A.

b)What is the domain of the areafunction?

c) For what value of x is the areaa maximum?

x

y

B A (x, y)

C D

Solution (b)

Domain of the area is 0 2x

Question 10 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola

y = 4 – x2 that is above the x-axis.

a)Express the area A of the rectangle as a function of the x-coordinate of A.

b)What is the domain of the areafunction?

c) For what value of x is the areaa maximum?

x

y

B A (x, y)

C D

Solution (c)

From the expression in (a),

when x 1.155, P(x) has a maximum.max (1.155) 6.158P

2( ) 2 (4 )A x x x

Assignment

Day 1P. 150 #16 – 24 (even),P. 161 #2, 3, 5, 7, 8, 10, 11

Day 2P. 162 #14, 15, 18, 19, 21, WS 4.3B