Largest volume of an open box.

Question

I need a quick check on this problem. Am I doing this right?

? 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.

So I setup my problem like this:

$$ x^2 + 4xh = 1200 = P$$

$$x^2 \cdot h = V$$

So I am now trying to express h in terms of x so I can get the volume equation down to 1 variable and then find the derivative of the volume equation. Is this approach right?

$$4xh = 1200 - x^2$$

$$\frac{1200 - x^2}{4x} = h$$

So,

$$ x^2 \cdot \frac{1200 - x^2}{4x} = V$$

$$ = \frac{x \cdot (1200 - x^2)}{4} = V$$

$$\frac{1}{4} ( \frac{d}{dx} \cdot ( x ( 1200 - x^2))) = V'$$

$$ = \frac{-2x^2 + 1200 - x^2}{4}$$ $$ = 300 - \frac{3x^2}{4}$$

$$V' = 0 $$ when x = 20

So max volume is $20 \cdot h$ and $h = \frac{1200-x^2}{4x} = 10$

So max volume = 300

Is this right?

Answer 1

Max vol is $x^2h$ not $xh$, hence $20^2\cdot 10=4000$, not $20\cdot 10$ (which is, by the way, 200, not 300).

Answer 2

Your solution is not correct (volume is $20^2h$ but you have calculated $20h$ {which is not even 300}) but formally you should have used double derivative test, the points at which the derivative is zero are called critical points and double derivative test is used to confirm if a critical point is a point at which the function attains maximum value.

Answer 3

Oh, no, I forgot how to use derivatives! Maybe I can solve this anyway, let's see.
$$ 1200=x^2+4xh=x^2+2xh+2xh\stackrel{\text{AM-GM}}{\geq} 3\sqrt[3]{x^2\cdot 2xh\cdot 2xh} =3\cdot4^{1/3} V^{2/3}$$ so $V\leq 4000$ and equality is attained iff $x^2=2xh$, i.e. iff $x=2h$, i.e. iff $(x,h)=(20,10)$.
Phew.