Maximum volume of a cuboid

Question

Of all the parallelopipeds whose area is equal to 2a find the one with the maximum volume?

I think this is a optimization problem, and I need to use differentiation to solve it.

Answer 1

For a parallelopiped with sides $A $, $ B $ and $ C $, the surface area will be given by $2 (AB + AC+BC) $, which yields $ a=AB+AC+BC $. Volume $ V $ of a parallelopiped is given by $ ABC $. Solving for $ C $ from the previous expression and substituting the rusult into the expression for volume, we get $ V (A, B)=\frac {AB (a-AB)}{A+B} $. Now all you need is to use differentiation.

Answer 2

You can also solve the problem without using any calculus. For any 3 vectors $\vec{x}, \vec{y}, \vec{z}$, we have:

$$( \vec{y} \times \vec{z} ) \times (\vec{z} \times \vec{x} ) = \left[ ( \vec{y} \times \vec{z} ) \cdot \vec{x} \right] \vec{z} - \left[ ( \vec{y} \times \vec{z} ) \cdot \vec{z} \right] \vec{x} = \left[ ( \vec{y} \times \vec{z} ) \cdot \vec{x} \right] \vec{z} $$ This implies $$ |\vec{x} \cdot ( \vec{y} \times \vec{z} )|^2 = (\vec{x} \times \vec{y} ) \cdot \left[ ( \vec{y} \times \vec{z} ) \times (\vec{z} \times \vec{x} ) \right]\tag{*} $$ and hence $$ |\vec{x} \cdot ( \vec{y} \times \vec{z} )|^2 \le |\vec{x} \times \vec{y}||\vec{y} \times \vec{z}||\vec{z} \times \vec{x}| \le \left(\frac{ |\vec{x} \times \vec{y}| + |\vec{y} \times \vec{z}| + |\vec{z} \times \vec{x}| }{3}\right)^3 $$ Apply this to the parallelepiped spanned by $\vec{x}, \vec{y}, \vec{z}$ with total surface area $2a$ and volume $V$, we get:

$$V \le \left(\frac{a}{3}\right)^{\frac32}$$ It is clear the equality can be achieved by a cube. In fact, if one reverse the logical deductions back to the step $(*)$, one will notice the equality will be achieved if and only if the 3 vectors $$\vec{x} \times \vec{y},\quad \vec{y} \times \vec{z}\quad\text{ and }\quad\vec{z} \times \vec{x}$$ are perpendicular to each other and has same magnitude. Geometrically, this means all six faces of the parallelepiped has same area and the dihedral angle between any two nearby faces are at $90^{\circ}$. This implies the equality is achieved when and only when the parallelepiped is a cube.