Calculus 1 - Optimization of a Box Can you guys help me out with it? i try to solve it but my answer is so weird that i think im wrong...
Question- Someone want to build cardboard box with rectangular base. Knowing thatthe rectangle of the base (and therefore the cover) is such that its width is twice its length, determine an dimensions of greater volume that can be built using 2 square meters of cardboard.
I frist make sure what i need to optimize, the volume. Then i try to get "h" by taking the total area of the box, putting that "h" on volume formula and deriving it i would get a nice result, that doesn't happened...
 A: From the conditions we know that $b=2a$ and we want to maximize $V = 2a^2c$ under the constraint that $2a^2 + 3ac = 1$. You can subsitute $c= \frac{1-2a^2}{3a}$ and you will get that:
$$V = \frac{2a^2(1-2a^2)}{3a} = \frac{2}{3}a(1-2a^2)$$
To find the extreme points take the derivative and you equal it to zero to get:
$$\frac 23 (1-6a^2) = 0 \implies a= \frac{1}{\sqrt{6}}$$
Therefore you can derive that $b=\frac{2}{\sqrt{6}}$, $c=\frac{1 - \frac 13}{\frac{3}{\sqrt{6}}} = \frac{2\sqrt{6}}{9}$ and $V=\frac{2\sqrt{6}}{27}$
A: OK the first step is to identify all the equations that are involved in the problem:


*

*W = 2L

*V = W * L * H

*A = 2 * W * L + 2 * W * H + 2 * L * H = 2
We can replace the equation 1 in equation 3 and then find the value of H in terms of L:
 2 = 4 * L² + 4 * L * H + 2 * L * H , 
 2 = 4 * L² + 6 * L * H , 
 H = (2 - 4*L²)/(6*L)
Now, we replace H and W in V:
V = 2 * L² * (2 - 4* L²)/(6 * L)
Our last step is find the derivative of V respect to L: 
V' = 2/3 - 4 * L²
Now we find the critical points which the first derivative is equal to 0. We can see that the second derivative is always negative to positive values of L, so we can establish that the positive value of the critical point L will be a maximum in V.
2/3 - 4* L² = 0
Then,
L = 1 / √6
That is the value of L that makes the maximum volume of the box, you can replace this value in the others equations and find the value of W,H,A and V
Have a nice day!
A: First I will recast your question into my interpretation:

Suppose we have 2 square meters of cardboard, and we wish to build a
  box (with no top) using the cardboard such that the width of the base of the box is
  twice the length of the base. How large can we build the box?

I will solve this in the general case where we have $\gamma$ square meters of cardboard. Let $V$ be the volume of the box, which is the product of the base length $\ell$ and width $w$ with the height $h$. Using the given constraints, we can set the volume $V$ in terms of  only variable $\ell$. We eliminite $w$ from the volume formula using the assumption $w=2\ell$. Next we eliminate $h$ using geometry. To make a box out of a piece of cardboard, four equal squares must be cut out from the corners of the cardboard, and then the shortened sides (making a cross) are folded up. The height $h$ is then the length of a side of one of the cut out squares. Then we see that the length and width of the piece of carbdoard may be written as $2h+\ell$ and $2h+w$ respectively (a picture makes this clear). The product of the length and width of the cardboard must be the assumed area of the cardboard, so we have
$$ \gamma = (2h+\ell)(2h+w) $$
which we simplify by substutution and expansion to
$$ \gamma = 4h^2 +6\ell h + 2 \ell^2 $$
Now solving for $h$ yields
$$ h = \pm \bigg({1 \over 2} \sqrt{-2\ell^2-2\ell \gamma + \gamma}\bigg) $$
and we take the positive value. Overall, we now have
$$ V = \ell h w = {2\ell^2\sqrt{-2\ell^2-2\ell \gamma + \gamma} \over 2} $$
which is what we must maximize. The derivative with respect to $\ell$ is
$$ {d \over d\ell} \bigg( \ell^2\sqrt{-2\ell^2-2\ell \gamma + \gamma} \bigg)
   = {\ell(-6\ell^2-5\ell\gamma+2\gamma)
     \over \sqrt{-2\ell^2-2\ell\gamma+\gamma}} $$
The denominator is zero when
$$ \ell={-\gamma \pm\sqrt{\gamma(\gamma+2)} \over 2} $$
so we must avoid these points. The numerator is zero when
$$ \ell={-5\gamma \pm\sqrt{25\gamma^2+48\gamma} \over 12} $$
Since the denominator is positive definite, we remove the negitive definite solution, so we are left with
$$ \ell={-5\gamma + \sqrt{25\gamma^2+48\gamma} \over 12} $$
and such $\ell$ will be the length when volume is maximized.
Substituting $\gamma=2$, the length when volume is maximized is then
$$ \ell={-5\cdot 2 + \sqrt{25\cdot 2^2+48 \cdot 2} \over 12} = {1 \over 3} $$
Substituting this value of $\ell$ (and $\gamma=2$) into the equation for $V$ in terms of $\ell$ above, we obtain
$$ V = {2\cdot{1 \over 3}^2\sqrt{-2 \cdot {1 \over 3}^2
   -2\cdot{1 \over 3} \cdot 2 + 2} \over 2} = {2 \over 27} $$
Modulo arithmetic errors, this is the maximum volume of the box with the given constraints.
Note: the other answers here assume that the initial piece of cardboard may by transformed non-isometrically (stretching, shrinking, etc.).
