1. ## Optimization Problem

If 1200 sq. cm of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

I understand the basics of how these problems work, but applying that is difficult. Do I start with V= xy or something similar?

2. Originally Posted by bobsanchez
If 1200 sq. cm of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

I understand the basics of how these problems work, but applying that is difficult. Do I start with V= xy or something similar?
The volume of the box is the area of the base times the height: $V=x^2y$.

Note that since the box has an open top, the amount of material is the area of the 5 remaining sides: $1200=4xy+x^2$

Can you solve this problem now?

3. Um, I think so. Setting it up is definitely the hardest part for me. So from there I try to put it in terms of y and then differentiate, correct? Should that be y' = -(4x^2 - 4800)/(4x)^2 ?

4. Originally Posted by bobsanchez
Um, I think so. Setting it up is definitely the hardest part for me. So from there I try to put it in terms of y and then differentiate, correct? Should that be y' = -(4x^2 - 4800)/(4x)^2 ?
You need to plug y back into the volume equation, and then differentiate!!

From the constraint, we see that $y=\dfrac{1200-x^2}{4x}$, which in return gives us the volume equation $V=x^2\cdot\dfrac{1200-x^2}{4x}=300x-\tfrac{1}{4}x^3$.

Thus, $V^{\prime}=\ldots$. Set $V^{\prime}=0$ and solve for x.

Can you proceed?

5. Ahhh that's what I've been doing wrong, okay. That makes much more sense. Let's see...I get V' = 300 - 3x^2/4, which when set equal to zero gives me x = 20...is this correct?

6. Originally Posted by bobsanchez
Ahhh that's what I've been doing wrong, okay. That makes much more sense. Let's see...I get V' = 300 - 3x^2/4, which when set equal to zero gives me x = 20...is this correct?
Correct!

Now what is y?

7. y still equals 300x - x^3/4.

So... V = 4000?