how do i solve it

Question

A rectangular box with a volume of 320ft^3 is to be constructed with a square base and top. The cost per square foot for the bottom is 15 cents for the top is 10 cents and for the sides is 2.5 cents. What dimensions will minimize the cost?

Answer 1

The volume of a box is lwh. Since we have a square base and top, the l and w are the same. Let x be the edges of the square, so the volume is x²h. 

 

So 320=x²h. 

 

The bottom has an area of x² and will cost 10c/ft². The top has an area of x² and will cost 15c/ft². Each side has an area of xh and will cost 2.5/ft². There is one top, one bottom, and four sides, so the equation for the total cost of the box is 

 

10x²+15x²+(4)(2.5)xh.

25x² + 10xh. This is the equation we wish to minimize. 

 

First we need to get the equation in terms of one variable. To do this, solve the volume equation for h.

 

So h=320/x²

 

So we can write the second equation as C(x)=25x²+3200/x. We want to figure out the minimum of this equation, so we take the derivative, set it equation to 0, and solve for x.

 

C'(x)=50x-3200/x²=0

So 50x=3200/x²

So 50x³=3200

So x³=64

So x=4 will minimize the cost function. We still need to find h. Recall that h=320/x². So then h=320/16.

 

Hopefully that was clear! I'm available for online tutoring if you would like more clarification.

 

Lena

Answer 2

Hi Lowe,

 

Here are a few questions to ask yourself to start off:

 

What does the word "minimize" tell you in the problem statement? How is minimization related to what you've been learning in Calculus?

 

How would you draw the box and label the variables? Are there any equations you could construct to get you started?

 

Tom

Answer 3

We are trying to minimize a cost function, so we need to write one.  The slope of the minimum cost will be 0, so differentiating the cost function and setting the first derivative equal to 0 should help.

 

When working optimization word problems, the trick is in the set-up, clearly.

 

So what do we know?

We know what the volume should be, and that the base and top should be square (of equal length).

How can we model that?
The volume of a rectangular box is V = L*W*H, where here L = W.  So then our volume formula becomes:

V = (L^2)*H.

 

Now, the cost function.  A box has six faces: a top, bottom, and four sides of equal area. We have been given costs associated for the material of each of these sides, so we should be able to write a cost function using the costs (given in cents per square foot):

C(L,H) = 15*L^2 + 10*L^2 + 4*2.5*L*H

The most important part of this problem is right here - you must understand how this cost equation was obtained.

Cost is in cents, and the 15, 10, and 2.5 are in cents/sq.ft, which are multiplied by dimensions L^2 or L*H.  Thus, each term on the right side also has units of cents.

 

Now we have a cost function, and we would desperately like to just differentiate this function, set it equal to 0, and then test the result to make sure that what we obtained was indeed a minimum.  However, this is Calculus I, not Calculus III, so we need to turn our cost function into one including only one variable.

 

The volume equation is the link - we can write the height in terms of the length and input this information into the  cost function.