Maximizing volume of a rectangular solid, given surface area

Solution 1:

Suppose that the sides of the faces that are guaranteed to be square is $x$, and the other $4$ sides are each equal to $y$. Then the sum of the areas of the faces is $2x^2+4xy$. We are told this is $6a^2$. Thus $$2x^2+4xy=6a^2.\tag{$1$}$$ We want to maximize the volume, which is $x^2y$.

Note that from Equation $(1)$ we obtain $y=\frac{3a^2-x^2}{2x}$. So, in terms of $x$, the volume is $x^2\cdot \frac{3a^2-x^2}{2x}$, which simplifies to $$\frac{3a^2x-x^3}{2}.$$ Call this function $V(x)$. Use the derivative to maximize $V(x)$. It turns out that the maximum is attained at $3a^2-3x^2= 0$, that is, when $x=a$. Now calculate the corresponding $y$. You will not be surprised to find that $y=a$, that is, that the optimal shape is a cube.

Remark: More generally, we may want to show that to maximize volume, we should make all sides equal, without being given the information that at least one face is square. There are several ways to do this. We could use tools from the calculus of several variables. But those are presumably at this point not available to you.

Or else we could let the sides be $x$, $y$, and $z$. Then $2xy+2yz+2xz=6a^2$. (i) Keeping $y$ fixed, use the calculus to show that for maximum volume we must have $x=z$. (ii) Now allow $y$ to vary, and use exactly the argument we gave above to conclude that $x=y=a$.

But there is an interesting purely algebraic way to handle the problem, the famous Arithmetic Mean/Geometric Mean Inequality, which has a purely algebraic proof. AM-GM (in the case of $3$ variables) says that if $p$, $q$, and $r$ are positive, then $$\frac{p+q+r}{3} \ge \sqrt[3]{pqr},$$ with equality precisely when $p=q=r$. To apply this result, let $p=xy$, $q=yz$, and $r=zx$. We know that $2xy+2yz+2zx=6a^2$, so $\frac{p+q+r}{3}=a^2$. Using AM-GM, we find that $$a^2 \ge \sqrt[3]{pqr}=\sqrt{3}{x^2y^2z^2}=(xyz)^{2/3},$$ with equality only when $x=y=z$. This says that $xyz \le a^3$, with equality only when $x=y=z$. So the volume is always $\le a^3$, and it is $a^3$ only when the three sides are equal.

Solution 2:

Hint: If your block is $b \times c \times f$, the volume is $V=bcf$ and the area is $A=6a^2=2(bc+cf+bf)$ We can use the given value for the area to eliminate one of $b,c,d$ so $f=\frac {6a^2-bc}{b+c}$, giving $V=bc\frac {6a^2-bc}{b+c}$. Now you can take $\frac {dV}{db}$ and $\frac {dV}{dc}$, and set them to zero, and get two equations for $b,c$. You should find that the block is a cube, so $b=c=f=?$.