# 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 $\displaystyle 600 in^2$. 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?

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.