Maximum volume of a box given perimeter and surface area What would be maximum volume of a rectangular box with a given perimeter $P$ and surface area $S$?
I tried to solve following equations, where $l$ is length, $b$ is base, $h$ is height, $P$ is the perimeter, $V$ is volume, and $S$ is the surface area.
$$2(l + b + h) = P  $$
$$2(lb + bh + hl) = S$$  
Need to maximize $$V=lbh$$.
Thanks.
 A: Hint
From the perimeter, you can extract $l$. From the surface, you can extract $b$. So, now, $V$ is only a function of $h$ and you want to maximize $V$. Then, ... 
I am sure that you can take from here.
A: I'll show that when volume is maximized, some two of the dimensions have to be equal.  Then you can assume WLOG that $b=h$, and the rest is algebra.
First note that
$$ l^3+b^3+h^3 = \underbrace{(l+b+h)^3 - 3(l+b+h)(lb+bh+hl)}_{\text{constant}} + 3lbh $$
Thus maximizing the product of the dimensions is the same as maximizing the sum of their cubes, that is, our problem is equivalent to this one:

Maximize $l^3+b^3+h^3$ subject to the constraints $l+b+h=\tfrac12 P$ and $lb+bh+hl=\tfrac12 S$. 

Next, note that
$$ (l+b+h)^2 = (l^2+b^2+h^2) + 2(lb+bh+hl) $$
so fixing the values of two of the parenthesized expressions actually fixes all three.  Thus our problem is equivalent to this one:

Maximize $l^3+b^3+h^3$ subject to the constraints $l+b+h=\frac12 P$ and $l^2+b^2+h^2 = \frac14 P^2-S$.

Now Lagrange multipliers gives the condition
$$ \exists \lambda_1,\lambda_2 :
\left[\begin{matrix} l^2 \\ b^2 \\ h^2 \end{matrix}\right]
= \lambda_1 \left[\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}\right]
+ \lambda_2 \left[\begin{matrix} l \\ b \\ h \end{matrix}\right] $$
This condition shows that the three vectors in it are linearly dependent, and so
$$ \left|\begin{matrix}
1 & l & l^2 \\
1 & b & b^2 \\
1 & h & h^2
\end{matrix}\right| = 0 $$
That's a Vandermonde determinant, so it being zero implies that some two of the variables are equal, as claimed.
A: $$l=P/2-b-h$$
$$(P/2-b-h)b+bh+h(P/2-b-h)=\frac{S}{2}$$
$$ Pb-b^2-bh+bh+hP-bh-h^2=S/2 $$
$$ Pb/2-b^2+hP/2-bh-h^2=S/2$$
$$  h^2+(b-P/2)h+(b^2-Pb/2+S/2)=0 $$
Solving this equation gives you $h$ with respect to $b$. 
$$ h_{1,2}=\dfrac{(P/2-b)\pm\sqrt{(b-P/2)^2-4(b^2-Pb/2+S)}}{2} =f_1(b) $$
$$l=P/2-b-h=f_2(b)$$
And now we have:
$$V=bf_1(b)f_2(b)=V(b)$$
Now apply derivation and determine min and max of this function...
A: The maximum value of $ab$ with condition $a+b=\text{constant}$ is given by $a=b=\text{constant}/2$...
Similarly in this case if we have $l=b=h=\text{perimeter}/6$, we will get the maximum volume for the same perimeter.
