Problem:A rectangular box is to be open-topped with a volume of $\displaystyle 600 in^2$. The material for its bottom costs $$\displaystyle 6/in^2$, and the material for its four sides costs $$\displaystyle 5/in^2$. Find the dimensions of the box that minimize the total cost of the material needed to construct the box.

My Solution:Let $\displaystyle h$ = height, $\displaystyle l$ = length, $\displaystyle w$ = width. Remember that the volume of a rectangular prism is $\displaystyle V = lwh$. We can solve for $\displaystyle h = \dfrac{V}{lw} = \dfrac{600}{lw}$.

Now we can write our equation for the total cost as:

$\displaystyle C = ($$$\displaystyle 6/in^2)lw + ($$$\displaystyle 5/in^2)2wh + ($$$\displaystyle 5/in^2)2lh$

$\displaystyle = 6lw + 10wh + 10lh$

$\displaystyle = 6lw + 10w\dfrac{600}{lw} + 10l\dfrac{600}{lw}$

$\displaystyle = 6lw + \dfrac{6000}{l} + \dfrac{6000}{w}$. (1)

We can take the derivative with respect to $\displaystyle l$:

$\displaystyle \dfrac{\partial C}{\partial l} = 6w - \dfrac{6000}{l^2} + 0 = 0$.

so $\displaystyle 6w = \dfrac{6000}{l^2}$ which gives us $\displaystyle w = \dfrac{1000}{l^2}$.

so we can substitute $\displaystyle w$ back into (1) and we have:

$\displaystyle = 6l\dfrac{1000}{l^2} + \dfrac{6000}{l} + \dfrac{6000}{\dfrac{1000}{l^2}}$

$\displaystyle = \dfrac{6000}{l} + \dfrac{6000}{l} + 6l^2$

$\displaystyle = \dfrac{12000}{l} + 6l^2$.

Once again we can take the derivative with respect to $\displaystyle l$:

$\displaystyle \dfrac{\partial C}{\partial l} = \dfrac{-12000}{l^2} + 12l = 0$.

so $\displaystyle 12l = \dfrac{12000}{l^2}$ which gives us $\displaystyle l^3 = 1000$ and thus $\displaystyle l = +10$ or $\displaystyle -10$, however $\displaystyle l$ must be positive so $\displaystyle l=10$.

Therefore we have: $\displaystyle l=10$, $\displaystyle w = \dfrac{1000}{l^2} = \dfrac{1000}{100} = 10$, and $\displaystyle h = \dfrac{600}{lw} = \dfrac{600}{100} = 6$.

Is that correct? Does that make sense?