# Optimization problem

I have a question on a homework assigment that says the following:

Design a rectangular box of width w, length l, and height h, which will hold 512 cubic centimeters. The sides of the box cost \$0.01/square centimeter and the top and bottom cost \$0.02/square centimeter. Find the dimensions of the box that minimize the total cost of materials used. Round these values to the nearest hundreth of a cm. Also show the second partials test to verify that you obtained a minimum, and state the actual minimum cost.

So, what I did was determine my minimizing function and my constraint. I have my minimizing function as V=lwh and my constraint as .04l + .04h + 2w = 512. I then find when my first partials are equal to 0 to find critical points. Then on that critical point, I use the second partials test to see if it was a minimum, but it wasn't. When I use the second partials test, I keep ending up with it telling me I found a maximum. I need some help into why I am coming up with a maximum result instead of a minimum result.

Or is my minimizing function wrong? Im wondering maybe I should minimize V = l + w + h which yields the perimeters instead of V=lwh which yields volume? That just occurred to me now...because I don't neccessarily want to minimize volume.

Aaliyah
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Commented:
Aaliyah9152,
You have been honest and said this is homework, I will not do it but will point you in the right direction (doing hoemwork is against the membership agreement). You have formulated the problem in correctly

Firstly find out the thing you need to minimise ie cost of material which is surface area related

area top and bottom  =2wl
area sides                =2hl+2wh

So cost      C = 0.2*2wl + 0.1*(2hl+2wh)

but the volume must be 512 so the wlh= 512

ie the problem is to minimise C

C = 0.2*2wl + 0.1*(2hl+2wh)

subject to     wlh= 512

There are a number of ways of doing this but it looks if you are just looking at the partials (ie no langrange multipliers). The first thing is to eliminate one of the varaiblee from the cost equation using the conatraint, say substitute w=512/lh. Now minimise the cost, ie calculate dC/dl  and  dC/dh  and equate to zero. You can do the rest

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Author Commented:
GwynforWeb,
Thanks for your response. And I was not looking for an answer =) I just seemed to have my constraint and what I actually wanted to minimize backwards. I started realizing that. And you are correct, we are not using Lagrange multipliers in this problem. Just setting the first partials equal to 0, find the critical points and then use the second partials test to verify we got a minimum =) But thank you for clarifying I was minimizing the wrong function!

Thanks again,
Aaliyah
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Commented:
I do not mind giving a push in the right direction, with out the correct function you would have gone round in circles for ever. Of course now you have minimise it! :-)
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Author Commented:
Ugh, I got stuck again :(  I'm sorry...

Well I got to the point where I set my first partial derivatives equal to zero, but I'm having a hard time actually coming up with my critical points.

I have the first partials as the following:
dC/dl = -10.24 l^(-2) + .04w
dC/dw  = -10.24w^(-2) + .04l

I solved the constraint as h=512/lw. I know what I got to do, but I'm getting confused when I set these first partials to zero. Usually, from the examples in class, we have factored, but that doesn't seem to be useful here =/ I'm confused in what method I should use to determine the critical points when the first partials are equal to zero. Please help again if you can...I will post a new question and increase the points if you want me to.

Aaliyah
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Author Commented:
Nevermind, I'm retarded =)

Too much calculus in one day fries your brain... =)
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Commented:
dC/dl    = -10.24 l^(-2) + .04w = 0
dC/dw  = -10.24w^(-2) + .04l  = 0

assuming your diffrentiation is correct then you must solve the above, clearly l=h=0 is a solution but there non-zero ones as well. Substitute for w into eqn 2 from eqn 1, terms will cancel and the equation is easily solved.
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Author Commented:
GwynforWeb,
Yea, I took a break from the problem and went back to it and it came to me. I had so many problems to do, I guess my brain was just fried =) I was so used to the problems in the book and I wanted to factor, but when I thought about it, I just solved on of the equations for w, and plugged that into the second equation. I got a final answer of: 6.35cm x 6.35 x 12.70cm will yeild a volume of 512 cubic cm and yeild a minimum total cost of \$4.84 / square centimeters. My answer makes sense, so I'm all set =)