How to find the largest volume of box with dimension limits The U.S. Postal Service carries small packages, but when the boxes get big, the post office folks take out a tape measure and measure the length x of the box, the width y of the box, and the height z of the box with all measurements in inches.
If
      x + 2 y + 2 z > 100 ,
then the box is rejected; otherwise it is accepted.
What measurements x , y and z give rise to the acceptable box with the biggest volume? 
 A: Constraint equation: $x+2y+2z = 100$, in order to obtain maximum volume. Also $V=x \cdot y \cdot z$. Solving for $x$ we get $x=100-2y-2z$, substitute into our volume equation to obtain $V=(100-2y-2z) \cdot y \cdot z$.
$V_y=100z-4yz-2z^2$
$V_z=100y-2y^2-4yz$.
Now we want to solve $V_y=0$ and $V_z=0$. Hence,
$100z-4yz-2z^2=0$ ... (1)
$100y-2y^2-4yz=0$ ... (2)
From equation (2) we get $y=50-2z$, substituting this into equation (1) we get $z=\frac{50}{3}$ and from the comments, due to symmetry, we get $y=\frac{50}{3}$. We can now solve for $x$ to get $x=\frac{100}{3}$.
Therefore, Length=$\frac{100}{3}$in, Width=$\frac{50}{3}$in, Height=$\frac{50}{3}$in.
A: Well, the volume of a box is given by:
$$\mathscr{V}_{\space\text{box}}=\text{L}_{\space\text{box}}\cdot\text{H}_{\space\text{box}}\cdot\text{W}_{\space\text{box}}\tag1$$
For some number $\text{n}$ it is rejected, and that happends when:
$$\text{L}_{\space\text{box}}+2\cdot\text{H}_{\space\text{box}}+2\cdot\text{W}_{\space\text{box}}\space>\space\text{n}\tag2$$
Now, what happends when we set it equal to $\text{n}$:
$$\text{L}_{\space\text{box}}+2\cdot\text{H}_{\space\text{box}}+2\cdot\text{W}_{\space\text{box}}=\text{n}\space\Longleftrightarrow\space\text{L}_{\space\text{box}}=\text{n}-2\cdot\text{H}_{\space\text{box}}-2\cdot\text{W}_{\space\text{box}}\tag3$$
We get for the volume:
$$\mathscr{V}_{\space\text{box}}=\left(\text{n}-2\cdot\text{H}_{\space\text{box}}-2\cdot\text{W}_{\space\text{box}}\right)\cdot\text{H}_{\space\text{box}}\cdot\text{W}_{\space\text{box}}=$$
$$\text{n}\cdot\text{H}_{\space\text{box}}\cdot\text{W}_{\space\text{box}}-2\cdot\text{H}_{\space\text{box}}^2\cdot\text{W}_{\space\text{box}}-2\cdot\text{W}_{\space\text{box}}^2\cdot\text{H}_{\space\text{box}}\tag4$$
Now, we need to solve:
$$
\begin{cases}
\frac{\partial\mathscr{V}_{\space\text{box}}}{\partial\text{H}_{\space\text{box}}}=0\\
\\
\frac{\partial\mathscr{V}_{\space\text{box}}}{\partial\text{W}_{\space\text{box}}}=0
\end{cases}\space\space\space\Longleftrightarrow\space\space\space\begin{cases}\text{W}_{\space\text{box}}\cdot\left(\text{n}-4\cdot\text{H}_{\space\text{box}}\right)-2\cdot\text{W}_{\space\text{box}}^2=0\\
\\
\text{H}_{\space\text{box}}\cdot\left(\text{n}-2\cdot\text{H}_{\space\text{box}}\right)-4\cdot\text{H}_{\space\text{box}}\cdot\text{W}_{\space\text{box}}=0
\end{cases}\tag5
$$
A: This problem has a symmetry that can be fruitfully exploited. Replacing $x$ by $2x'$, we have for the dimension constraint $2(x'+y+z)=100$ and $2x'yz$ for the volume of the box. Because of the symmetry of the roles of the three variables, something interesting is bound to happen when $x'=y=z$. Indeed, the AM-GM inequality tells us that the volume of the box is maximized when this occurs. Substituting back into the dimension constraint yields $x=\frac{100}3$ and $y=z=\frac{50}3$.
