Q.If the sum of the surface areas of a cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere when the sum of their volumes is minimum?

Concept: Optimization Word Problem — using a constraint to express one variable in terms of the other, then minimizing the sum of volumes.

Let the edge of the cube be $a$ and the radius of the sphere be $r$. The constant sum of surface areas is:

We need to minimize the sum of volumes:

From the constraint, $a^2 = \frac{k - 4\pi r^2}{6}$. Differentiate $V$ with respect to $r$ (or use the method of Lagrange multipliers). Setting $\frac{dV}{dr} = 0$ gives:

From the constraint, $12a \frac{da}{dr} + 8\pi r = 0 \implies \frac{da}{dr} = -\frac{2\pi r}{3a}$. Substituting and simplifying yields: …

This is a classic optimization problem from calculus, but the real insight is geometric: both shapes have surface area proportional to the square of a linear dimension, and volume proportional to the cube. When you fix total surface area, you’re trading off between two “square” costs to minimize a “cubic” sum. The minimum occurs where the marginal volume gain per unit surface area is equal for both shapes — a condition that leads to a surprisingly clean ratio.

Let’s set it up.

1. Define variables and the constraint

Let the cube have edge length $a$, and the sphere have radius $r$.

Surface area of cube: $S_c = 6a^2$

Surface area of sphere: $S_s = 4\pi r^2$

The total surface area is constant, say $K$:

We want to minimize the total volume:

2. Reduce to one variable

From the constraint, express $a^2$ in terms of $r$:

So $a = \sqrt{\frac{K - 4\pi r^2}{6}}$.

Then volume becomes a function of $r$ alone:

3. Differentiate and set to zero

We need $\frac{dV}{dr} = 0$. Differentiate term by term.

For the cube term: let $u = \frac{K - 4\pi r^2}{6}$, then $a = u^{1/2}$, so $a^3 = u^{3/2}$.

Now $\frac{du}{dr} = \frac{-8\pi r}{6} = -\frac{4\pi r}{3}$.

So derivative of cube volume:

For the sphere term:

Set sum to zero:

4. Solve for $r$

Factor out $2\pi r$ (note $r>0$ for a non-degenerate sphere):

So:

Square both sides:

Multiply through:

$K = 24r^2 + 4\pi r^2 = 4r^2(6 + \pi)$ …

Method: Optimizing a Sum of Two Shapes' Volumes Under a Shared Surface-Area Constraint

When two different shapes share a single constraint (their combined surface area is fixed) and you must optimize the sum of a different property (their combined volume), the general technique is to differentiate the constraint implicitly to link the two shapes' rates of change, then apply that link to the derivative of the objective.

Steps

Step 1: Assign a variable to each shape's defining length and write both formulas

Give each shape its own variable (e.g. cube edge $a$, sphere radius $r$). Write down the surface-area formula and the volume formula for each shape in terms of its own variable.

Step 2: Write the constraint and the objective

Add the two surface areas and set the sum equal to the given constant — this is the constraint linking $a$ and $r$. Add the two volumes to form the objective function to be minimized (or maximized).

Step 3: Differentiate the constraint implicitly, then the objective, and combine …

Common Mistakes

Mistake 1: Reporting the ratio in terms of radius instead of diameter

The problem specifically asks for edge : diameter, not edge : radius. Why it's wrong: the working naturally produces $a=2r$ (edge equals twice the radius), so stopping there and writing the ratio as $a:r=2:1$ answers the wrong question — the diameter is $2r$, making the correct ratio $a:2r=1:1$. Correct approach: convert the radius into a diameter ($d=2r$) as the very last step before stating the final ratio.

Mistake 2: A sign or chain-rule slip while implicitly differentiating $a$ with respect to $r$ …