Math 221 Week 9 part 2

More optimization word problems

Please take a moment to just breathe.

We solve two more optimization word problems.

General Method for solving optimization problems.

Step 1. Draw a picture and give things names.

Step 2. Which quantity do we want to optimize?

Step 3. What bounds are naturally given by the problem?

Step 4. What constraints are given? What relationships between quantities are given?

Step 5. Write the quantity you want to optimize in terms of just one other variable.

Step 6. Take the derivative and find the critical points.

Step 7. Use the first or second derivative test to see which critical point is optimal. Also check endpoints, if relevant.

Step 8. Do a reality check!

Example.

A hallway of width 5 ft meets a hallway of width 4 ft at a right angle. Find the length of the longest board than can be carried horizontally around the corner. (Assume you do not tilt the board vertically.)

Step 1. Draw a picture and give things names. Step 2. Which quantity do we want to optimize?

We want to find the length of the longest plank that fits.

4

5L

Let � be the angle shown.

Define � to be the length of the plank that just touches both walls and the corner as shown when the angle is � .

When � is very close to 0 or 90 degrees, then there’s plenty of room for the plank.

We want to minimize � as a function of � . Then any plank of length up to that minimum � will fit.

αL(α)

α

α

L αL

4

5

L�α

We split � up into two lengths: � , where

�

� or �

Together, we get �

L L = L1 + L2

sin α = 5/L2

sin(π/2 − α) = 4/L1 cos α = 4/L1

L =5

sin α+

4cos α

4

5

L1�αL2

Step 6. Find the critical points of the function � .

�

� �

Critical point: �

Step 7. We already know this value will be a minimum.

We compute �

The largest ladder that fits is about 12.7 feet long.

L(α)

L =5

sin α+

4cos α

dLdα

= −5 cos αsin2 α

+4 sin αcos2 α

=−5 cos3 α + 4 sin3 α

sin2 α cos2 α

α = arctan( 3 5/4) ≈ .82255

L(.82255) = 12.7

Example. Find the point on the parabola � closest to the point (3,9).

Steps 1 and 2. Draw a picture and give things names.

� = distance from (3,9) to (x,y)

= �

It’s easier to use � :

�

If � is minimized, so is � .

y = 9 − x2

D

(x − 3)2 + (y − 9)2

D2

D2 = (x − 3)2 + (y − 9)2

D2 D

(3,9)

(x,y)

D

Example. Find the point on the parabola � closest to the point (3,9).

We want to minimize �

�

subject to the constraint that

�

So we need to minimize

�

�

y = 9 − x2

D2

D2 = (x − 3)2 + (y − 9)2

y = 9 − x2

D2 = (x − 3)2 + ((9 − x2) − 9)2D2 = (x − 3)2 + x4

(3,9)

(x,y)

Now we need to find the critical points.

� , so �

We have to find the roots of this cubic.

The teacher tells us that one root is � .

We factor:

�

and notice that the second term has no other roots.

So the only critical point is � .

D2 = (x − 3)2 + x4 (D2)′� = 2x − 6 + 4x3

x = 1

4x3 + 2x − 6 = (x − 1)(2x2 + 2x + 3)

x = 1

Example. Find the point on the parabola � closest to the point (3,9).

We know that the only critical point is � .

We check the second derivative there:

�

�

�

� is concave up, so there’s a

minimum at our critical point.

Finally, we check that � makes sense.

y = 9 − x2

x = 1

D2 = (x − 3)2 + x4(D2)′� = 2x − 6 + 4x3(D2)′�′� = 2 + 12x2 > 0

D2

x = 1

(3,9)

(x,y)