Up: Notes for Math 15

# Notes on Optimization Problems:

A procedure for handling optimization problems is as follows:

Example: You build open-topped boxes and you want to keep your costs down. You need a box with volume 32 cubic units, and you need the base of the box to be at least 2 ´ 2 square units and the height to be at least 1 unit. Find the dimensions of the box that uses the least materials.

1. Rewrite the problem with one piece of information per line.
Rectangular boxes (no top)
Volume 32
Base at least 2 ´ 2
Height at least 1
Find dimensions
Use least material

2. Draw several pictures illustrating different possibilities.
Some possibilities include a 2 ´ 2 ´ 8 box, a 4 ´ 4 ´ 2 box, and a 4 ´ 8 ´ 1 box.

3. Draw a generic picture and introduce variables.
The width, depth and height are varying, so we let x be the width, y be the depth and z be the height of the box, and draw an x ´ y ´ z box.

4. Explicitly write down "Maximize ________" or "Minimize ________" and find a formula for the quantity to be optimized.
Minimize materials. (I.e., minimize the area of the materials used.)

Let A = area of materials.
Then A = xy + 2xz + 2yz is the function to be minimized.

5. Use information you know to eliminate all but one variable.
We know that the volume is 32. So xyz = 32 since the volume is width times depth times height. Solving for z we get z = 32/xy, which we can plug into the formula for area to eliminate one variable. So A = xy + 2x(32/xy) + 2y(32/xy), or

A(x,y) = xy + 64/y + 64/x

is the function to be minimized.

6. Determine the region R where the function makes sense.
We know the base must be at least 2 ´ 2, so x > 2 and y > 2. We also know that z > 1, so (32/xy) > 1, or 32 > xy (since x and y are both positive the inequality doesn't change). Since x > 2, we know xy > 2y, so 32 > xy > 2y, or 32 > 2y, which means y < 16. Similarly, x < 16. Thus the region is the square pictured.

7. Locate all critical points inside R (discard any outside R).
We need to find where ÑA = (0,0). First, we compute the gradient:

ÑA(x,y) = <y - 64/x2, x - 64/y2>

and this equals <0,0> when y - 64/x2 = 0 and x - 64/y2 = 0. That is, when y = 64/x2 and x = 64/y2. The latter becomes xy2 = 64, into which we substitute the former to get x(64/x2)2 = 64, which becomes 64 = x3, or x = 4. Since y = 64/x2, this means y = 4 as well.

Thus ÑA = <0,0> at (4,4) and no other point. This point is inside the region R, so it is one of our candidates for the minimum value of A.

8. Parameterize the boundary of R in terms of t (using several pieces if necessary).
Since R is a square, we can parameterize each edge separately. For example, the bottom edge goes through the point (2,2) and has direction vector <1,0>, so it can be parameterized by

L1(t) = (2,2) + t <1,0> = (2+t,2).

Similarly the other edges are

L2(t) = (2,2+t)
L3(t) = (2+t,16)
L4(t) = (16,2+t)

9. Form the composition of each boundary parameterization with the function to be optimized.
For line L1, we obtain

 F1(t) = A(g(t)) = A(2+t,2) = (2+t)(2) + 64/(2+t) + 84/2 = 2t + 64/(2+t) + 36.

Note that the interval of t values for this line is 0 < t < 14.

The other lines work similarly.

10. Optimize the compositions as function of one variable (i.e., find where F'(t) = 0 or F' is undefined).
For our F1, F1'(t) = 2 - 64/(2+t)2. This is undefined when (2+t)2 = 0; i.e., when t = -2. This is outside of our interval of t values, so we discard it.

To find where the derivative is 0, we solve for t in 2 - 64/(2+t)2 = 2, so (2+t)2 = 32, or t = sqrt(32) - 2, which is approximately 3.66. This produces the point L1(3.66) = (5.66,2) as a candidate for the minimum value of A, togehter with the two endpoints of the edge at (2,2) and (16,2).

The other three sides can be treated similarly.

11. Check the value of the function at each point obtained in parts 7 and 10. Find the largest or smallest of these. This is the optimal value.
We need to check the points (4,4), (5.66,2), (2,2), (2,16) and the other points obtained from the three other line segments that form the edges of R.
A(4,4) = (4)(4) + 64/4 + 64/4 = 16 + 16 + 16 = 48
A(5.66,2) = (5.66)(2) + 64/5.66 + 64/2 » 54.63
A(2,2) = (2)(2) + 64/2 + 64/2 = 4 + 32 + 32 = 68
A(2,16) = (2)(16) + 64/16 + 64/2 = 32 + 4 + 32 = 68
Notice that the smallest of these is the first one. If you do the other three edges of R you will find that they are larger than 48, so this is the minimum value of A.

12. Compute the quantity actually requested in the problem.

We are looking for the dimensions of the box that yield the minimum area, so we still need to find these. We already have x = 4 and y = 4, so we still need to get z. Since we found above that z = 32/xy, we can use this to find z = 32/(4×4) = 32/16 = 2. Thus the optimal box has a 4 ´ 4 base and is 2 units tall.

Up: Notes for Math 15