# 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.

• See this. Aug 3, 2014 at 0:56

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.

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.

$$l=P/2-b-h$$

$$(P/2-b-h)b+bh+h(P/2-b-h)=\frac{S}{2}$$ $$Pb-b^2-bh+bh+hP-bh-h^2=S/2$$ $$Pb/2-b^2+hP/2-bh-h^2=S/2$$ $$h^2+(b-P/2)h+(b^2-Pb/2+S/2)=0$$

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...

• First statement $l=P−b−h$ is false Apr 16, 2014 at 9:28

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.