# Thread: Find dimensions to minimize total cost of materials

1. ## Find dimensions to minimize total cost of materials

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

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

Now we can write our equation for the total cost as:
$C = ($$$6/in^2)lw + ($$ $5/in^2)2wh + ($\$ $5/in^2)2lh$

$= 6lw + 10wh + 10lh$

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

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

We can take the derivative with respect to $l$:
$\dfrac{\partial C}{\partial l} = 6w - \dfrac{6000}{l^2} + 0 = 0$.

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

so we can substitute $w$ back into (1) and we have:
$= 6l\dfrac{1000}{l^2} + \dfrac{6000}{l} + \dfrac{6000}{\dfrac{1000}{l^2}}$

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

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

Once again we can take the derivative with respect to $l$:
$\dfrac{\partial C}{\partial l} = \dfrac{-12000}{l^2} + 12l = 0$.

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

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

Is that correct? Does that make sense?

2. ## Re: Find dimensions to minimize total cost of materials

Hey vidomagru.

That is spot on: well done .

Intuitively your answer also makes sense since a square will provide the biggest area and/or volume under optimization if it has to be rectangular shape.