Question

A rectangular box with an open top is designed to have a volume of 128 cubic feet. The sides of the box cost twice as much per square foot as the bottom. Find the dimensions of the box for which the cost of all sides is a minimum.

Answer 1

how do i start this when there are three variables? i get v=b*h*l v= 128

Answer 2

i get 512+512dy/dx+dy/dx which is dy/dx = 512/513eventually but im unsure of the next step

Answer 3

Have you done Lagrange Multipliers?

Answer 4

If you accept symmetry arguments, this problem can be solved without Lagrange multiplier, just simple calculus.

Answer 5

@robertrico
Since you do not seem to be online, and I will probably be offline soon, here's my suggestion to the approach.

First, express the cost as a function of the three dimensions, x,y=dimensions in plan, and z=height.
We note that this is an open box.
Let p=unit price for the bottom, then 2p=unit price for the sides.
Total cost
C(x,y,z)=p(xy)+2p(2xz+2yz)
=p(xy)+4p(xz+yz)

Since a square is a quadrilateral with the least perimeter for a given area (can be proved using calculus), we assume x=y, which reduces the cost equation to:
\(C(x,z)=px^2+4pxz\)
Since we already know the volume, or
xyz=128, we conclude that
\(z=128/xy=128/x^2\) for x=y (assumed above)
Substituting z into the cost function, we reduce the cost function to a single variable x.
\(C(x,128/x^2)=px^2+\frac{1024p}{x}\)
or simply
\(C(x)=px^2+\frac{1024p}{x}\)
Now you can merrily go ahead and equate the derivative and solve for x.

Hint:
It should turn out to be a nice integer which is a factor of 37592.

Answer 6

*equate to zero