# Homework Help: Help with dreaded word problem

1. Aug 3, 2008

### illjazz

1. The problem statement, all variables and given/known data
If 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

2. Relevant equations
- Differentiation rules
- Magic?

3. The attempt at a solution
First, I drew a square. I drew little squares in the corners of the big square. The sides of the little squares I labeled y. Each side of the big square has a length of $$\sqrt{1200}$$. The length of the sides of the box that is to be built is $$\sqrt{1200} - 2y$$ and is labeled x.
The volume of the box is given by

$$V=yx^2$$

We know $$x=\sqrt{1200}-2y=20\sqrt{3}-2y$$

So we get a function

$$V(y)=y(20\sqrt{3}-2y)^2$$

The next step, I presume, is to take the derivative of this function. Then we have

$$V'(y)=4(3y^2-40y\sqrt{3}+300)$$

As a first step in finding a maximum for this function, I set V' equal to zero:

$$4(3y^2-40y\sqrt{3}+300)=0$$

I get

$$y=0$$ or $$y=10\sqrt{3}$$ by asking my calculator.

(Trying to calculate the "zeroes" for this equation by hand ends in disaster..

Here's my attempt:

$$12y^2-160y\sqrt{3}+1200=0$$

$$12y^2-160y\sqrt{3}=-1200$$

$$3y^2-40y\sqrt{3}=-300$$

$$y^2-\frac{40\sqrt{3}}{3}y=-100$$

Eh!?

How would I go from here to get the zeroes? That's just a side question..)

So, now I can make a number line, or a "wiggle graph":

$$V'(-1)=40(40\sqrt{3}+303)$$, which is positive.
$$V'(1)=-40(40\sqrt{3}+303)$$, which is positive.
$$V'(10\sqrt{3}+1)=40(20\sqrt{3}+3)$$, which is positive...

Code (Text):

---+-----------+--------------------+------
---------(0)---------(10\sqrt{3})-----------

Hmm.. they're not supposed to be all positive. How am I supposed to find a maximum if V' never goes from positive to negative? There is obviously something wrong here..

2. Aug 3, 2008

### Redbelly98

Staff Emeritus
I interpret that to mean the total surface area of the box (excluding the top, which is open) is 1200 cm^2.

3. Aug 3, 2008

### illjazz

Yes, but as I understand it, that does not change the volume of the box. The volume will be the same, regardless of whether there is a top, no?

4. Aug 3, 2008

### Redbelly98

Staff Emeritus
Correct. But it does change the relation between the side length and height (x and y).

5. Aug 3, 2008

### illjazz

How so? I don't think I'm following. I mean I see what you're saying, but I don't see how it affects the volume, which is what I'm after :/

Setting this issue aside, does my process otherwise look correct?

6. Aug 3, 2008

### Redbelly98

Staff Emeritus
The process is otherwise correct, yes. Set the derivative of V equal to zero to find the maximum.

V = x^2 y, as you said.
If you substitute a different expression for x (in terms of y) than the one you did, you'll get a different maximum volume.

7. Aug 3, 2008

### Redbelly98

Staff Emeritus
Are you familiar with the Quadratic Formula? And do you see why y=0 is clearly not a solution to this equation?

8. Aug 3, 2008

### illjazz

Even though we're not counting the "top" of the box, a side's length will still be
$$x=\sqrt{1200}-2y$$, will it not? If this is where my error lies, I do not see it.

I am, of course. I did have to raise my eyebrow when I calculated V'(0). I got 1200! Obviously can't be right.. that's what my calculator gave me for solving that equation though.

I do see why it isn't possible.. also, I'm stuck on the by-hand method of solving the equation for zero, as I mentioned. Arg.. you just addressed that by mentioning the quad formula! I'll give that a try.

Also, what's up with getting all positive values on the "wiggle graph"? I think the problem I'm having right now is understanding just what I'm missing. Since this is kind of a multilayered problem and I seem to have more than one mistake in there, I'm having trouble piecing it all together. If the foundation is unstable, the rest of the construct can't be faring too well...
PHP:

9. Aug 4, 2008

### Redbelly98

Staff Emeritus
The wiggle graph isn't reliable until you've found the zeroes, so I wouldn't worry about that just yet. I get a negative V' for y=6, by the way, so once you've gotten the zeroes from the quadratic formula it should work out.

You have assumed that the box is to be cut out of a 1200 cm^2 square piece of material, which I'm not certain is right. Another interpretation of the problem statement is: find the total surface area of the box in terms of x and y, and set that expression equal to 1200.

At the very least you'll get points for finding a function maximum using calculus, which is the real goal of this type of problem.

10. Aug 4, 2008

### HallsofIvy

You seem to be assuming that you start with a 1200 sq. cm. piece of material, cut out square corners and then turn up the sides to form a box. That is not the only way to make a box! Suppose you just cut out the bottom and four sides separately, then glued them together? That way, you don't throw away any material. Exactly what values x and h (height of the box) can be depend on the shape of the original piece which can be anything.

Suppose the square bottom has sides of length x and the height is h. Then the volume is V= x2h. The base has area x2 and the four sides each have area xh so the whole thing has area x2+ 4xh. If you cut the base and sides as efficiently as possible, x2+ 4xh= 1200. You can solve that for h and by replacing h in x2h by that, get V as a function of x only.