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

- Rewrite the problem with one piece of information per line.
Rectangular boxes (no top)

Volume 32

Base at least 2 x 2

Height at least 1

Find dimensions

Use least material

- Draw several pictures illustrating different possibilities.
Some possibilities include a

2 x 2 x 8 box, a4 x 4 x 2 bnox, and a4 x 8 x 1 box. - 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 box.*x*x*y*x*z* - Explicitly write down "Maximize ________" or "Minimize ________" and
find a formula for the quantity to be optimized.
Maximize materials. (I.e., minimize the area of the materials used.)

Let

*A*= area of materials.

Then*A*=*xy*+ 2*xz*+ 2*yz* - Use information you know to eliminate all but one variable.
We know that the volume is 32. So

since the volume is width times depth times height. Solving for*xyz*= 32*z*we get , which we can plug into the formula for area to eliminate one variable. So*z*= 32/*xy**A*= , or*xy*+ 2*x*(32/*xy*) + 2*y*(32/*xy*)*A*(*x*,*y*) =*xy*+ 64/*y*+ 64/*x*is the function to be minimized.

- Determine the region
*R*where the function makes sense.We know the base must be at least

2 x 2 , so and*x*__>__2 . We also know that*y*__>__2 , so*z*__>__1(32/ , or*xy*)__>__132 (since__>__*xy**x*and*y*are both positive the inequality doesn't change). Since , we know*x*__>__2 , so*xy*__>__2*y*32 , or__>__*xy*__>__2*y*32 , which means__>__2*y* . Similarly,*y*__<__16 . Thus the region is the square pictured.*x*__<__16

- Locate all critical points inside
*R*(discard any outside*R*).We need to find where

Grad( . First, we compute the gradient:*A*) = (0,0)Grad( *A*) (*x*,*y*) = (*y*- 64/*x*^{2},*x*- 64/*y*^{2})and this equals (0,0) when

and*y*- 64/*x*^{2}= 0 . That is, when*x*- 64/*y*^{2}= 0 and*y*= 64/*x*^{2} . The latter becomes*x*= 64/*y*^{2} , into which we substitute the former to get*xy*^{2}= 64 , which becomes*x*(64/*x*^{2})^{2}= 6464 = , or*x*^{3} . Since*x*= 4 , this means*y*= 64/*x*^{2} as well.*y*= 4Thus

Grad( at (4,4) and no other point. This point is inside the region*A*) = (0,0)*R*, so it is one of our candidates for the minimum value of*A*. - 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*L*_{1}(*t*) = (2,2) +*t*<1,0> = (2+*t*,2).Similarly the other edges are

*L*_{2}(*t*) = (2,2+*t*)

*L*_{3}(*t*) = (2+*t*,16)

*L*_{4}(*t*) = (16,2+*t*)

- Form the composition of each boundary parameterization with the
function to be optimized.
For line

*L*_{1}, we obtain*F*_{1}(*t*)= *A*(*g*(*t*))= *A*(2+*t*,2)= (2+ *t*)(2) + 64/(2+*t*) + 84/2= 2 .*t*+ 64/(2+*t*) + 36Note that the interval of

*t*values for this line is0 .__<__*t*__<__14The other lines work similarly.

- Optimize the compositions as function of one variable (i.e., find where
*F*'(*t*) = 0 or*F*' is undefined).For our

*F*_{1}, . This is undefined when (2+*F*_{1}'(*t*) = 2 - 64/(2+*t*)^{2}*t*)_{2}= 0; i.e., when . This is outside of our interval of*t*= -2*t*values, so we discard it.To find where the derivative is 0, we solve for

*t*in2 - 64/(2+ , so*t*)_{2}= 2(2+ , or*t*)_{2}= 32 , which is approximately 3.66. This produces the point*t*= sqrt(32) - 2*L*_{1}(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.

- 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 (about)

*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

*R*you will find that they are larger than 48, so this is the minimum value of*A*.Of course, we are looking for the

*dimensions*of the box that yield the minimum area, so we still need to find these. We already have and*x*= 4 , so we still need to get*y*= 4*z*. Since we found above that , we can use this to find*z*= 32/*xy* . Thus the optimal box has a*z*= 32/(4x4) = 32/16 = 24 x 4 base and is 2 units tall.

Comments to: dpvc@union.edu

Created: 2 March 1998 --- Last modified: Mar 2, 1998 1:51:50 PM