Dimensions for the cheapest possible rectangular box What are the dimensions for the cheapest possible rectangular box with a volume of 504 cm3  if the material for the bottom costs \$20/cm2, material for the sides costs \$6/cm2, and material for the top costs \$8/cm2 ?

I not sure how to use the information that I was given to solve this question. I know that I should use more 6 dollar pieces but I don't know how many of each I should use without doing guess and check.
For length $l$ cm, width $w$ cm and height $h$ cm, the material costs $c$ in dollars are
$c= 20lw+ 6(2lh+2wh)+ 8lw$ $=28lw + 12h(l+w)$
Note that the set volume of the box means that $h=504/lw$, so
$c=28lw + 6048(l+w)/lw$ $=28lw+6048/l + 6048/w$.

Then the partial derivatives on $l$ and $w$ are
$\frac{\delta c}{\delta l} = 28w-6048/l^2  $ and
$\frac{\delta c}{\delta w} = 28l-6048/w^2  $  

Solving for zero derivative we get 
$l^2w = 216$ and $w^2l= 216 $  

and thus $l=w$ at the minimum and we can solve for the dimensions.



Initial answer

It's relatively obvious that the top and bottom should be square, since we're looking to get minimum perimeter for the area.

So given that $s$ is the side length for the square base, the height $h(s)=504/s^2$ and the cost $c(s) = 20s^2+ 6\cdot 4sh(s)+ 8s^2$ $=  28s^2 + 12096/s$.

Then $\frac{dc}{ds} = 56s-12096/s^2$ and we can solve for the minimum at $\frac{dc}{ds} =0$.
For length $l$ cm, width $w$ cm and height $h$ cm, the material costs $c$ in dollars are
$c= 20lw+ 6(2lh+2wh)+ 8lw$ $=28lw + 12h(l+w)$
Note that the set volume of the box means that $h=504/lw$, so
$c=28lw + 6048(l+w)/lw$ $=28lw+6048/l + 6048/w$.

Then the partial derivatives on $l$ and $w$ are
$\frac{\delta c}{\delta l} = 28w-6048/l^2  $ and
$\frac{\delta c}{\delta w} = 28l-6048/w^2  $  

Solving for zero derivative we get 
$l^2w = 216$ and $w^2l= 216 $  

and thus $l=w$ at the minimum and we can solve for the dimensions.



Initial answer

It's relatively obvious that the top and bottom should be square, since we're looking to get minimum perimeter for the area.

So given that $s$ is the side length for the square base, the height $h(s)=504/s^2$ and the cost $c(s) = 20s^2+ 6\cdot 4sh(s)+ 8s^2$ $=  28s^2 + 12096/s$.

Then $\frac{dc}{ds} = 56s-12096/s^2$ and we can solve for the minimum at $\frac{dc}{ds} =0$.