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.
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.
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.