What are the dimensions for the cheapest possible rectangular box with a volume of $490 cm^3$ if the material for the bottom costs $\$6/cm^2$, material for the sides costs $\$7/cm^2$, and material for the top costs $\$14/cm^2$ ?
I can't seem to get anywhere close to the answer in $x,y,z$ form.
Volume = $x y z = 490$.
Cost = $\underbrace{6 x y}_{\rm bottom} + \underbrace{14 x y}_{\rm top} + 7 (\underbrace{2 x z}_{\rm front\ and\ back} + \underbrace{2 y z}_{\rm left\ and\ right}) = 20 x y + 14 z (x + y)$.
Now, minimize the cost, given the constraint of volume by taking the derivative of the cost with respect to $x$, setting it to zero. (The $z$ can be eliminated algebraically and we'll use the symmetry of $x$ and $y$.)
${\partial C \over \partial x} = {6860 \over x y} + 20 y - {6860 (x+y) \over x^2 y}$.
Set to zero and find $x = {7 \sqrt{7} \over \sqrt{y}}$.
Note that all the equations are symmetric with respect to the interchange $x \leftrightarrow y$. After all, there's no difference mathematically between left-right and front-back. So $x^{3/2} = 7 \sqrt{7}$:
Finally: $x = y = 7$ and $z = 10$.