1. ## optimization problem help please!!

Anything helps please. I don't know where to start besides creating a formula.
Here is the problem.

A rectangular box with a square base and top is to have a volume of 20 ft^3. The material for the bottom costs 30cents per square foot, the material for the sides cost 10cents per squaring foot, and the material for the top costs 20cents per square foot. What is the minimum cost of contructing such a box?

2. ## Re: optimization problem help please!!

Originally Posted by illicitkush
Anything helps please. I don't know where to start besides creating a formula.
Here is the problem.

A rectangular box with a square base and top is to have a volume of 20 ft^3. The material for the bottom costs 30cents per square foot, the material for the sides cost 10cents per squaring foot, and the material for the top costs 20cents per square foot. What is the minimum cost of contructing such a box?
let $\displaystyle x$ = side length of the square base and top

$\displaystyle h$ = box height

volume ... $\displaystyle x^2h = 20$

let $\displaystyle C$ = cost of the box in cents (if you want dollars, change to decimals)

$\displaystyle C = 30x^2 + 20x^2 + 10(4xh)$

solve for $\displaystyle h$ in the volume equation, substitute the result into the cost equation to get cost strictly as a function of $\displaystyle x$ ... minimize the cost like you were taught in class.

3. ## Re: optimization problem help please!!

When I solve for h do I leave C Inside the function or do I change it?

4. ## Re: optimization problem help please!!

When you solve for h in the equation concerning volume, substitute that into the cost function, and then as stated, you will have cost as a function of x alone. Then you may minimize this in the usual fashion.

5. ## Re: optimization problem help please!!

Okay so for h i got: h=20/x^2

Then I plugged that into Cost equation and got: C = 50x^2 + 800/x

Then I took the derivative and found x = 2.

So now I'm stick here?

I actually took x=2 plugged it back into the Cost equation C = 50x^2 + 800/x and got C=800, so I moved the decimal over twice and got $8 for the answer? 7. ## Re: optimization problem help please!! When you substitute for$\displaystyle h$you get:$\displaystyle C(x)=50x^2+40x\left(\frac{20}{x^2} \right)=50\left(x^2+16x^{-1} \right)$Differentiating, we find:$\displaystyle C'(x)=50\left(2x-16x^{-2} \right)=100\left(\frac{x^2-8}{x^3} \right)$Since we require$\displaystyle 0<x$, we find the critical value$\displaystyle x=2$, and the first derivative test verifies this is a minimum. Hence:$\displaystyle C_{\text{min}}=C(2)=600$and since this is in cents, we have a minimum cost of$6.

The only error you made was in evaluating C(2).

8. ## Re: optimization problem help please!!

Ohh oops I multiplied the 200 + 400. Got it
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