Q.The sum of the surface areas of a rectangular parallelopiped with sides $x$, $2x$ and $\dfrac{x}{3}$ and a sphere is given to be constant. Prove that the sum of their volumes is minimum if $x$ is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.

Concept: Optimization Word Problem — using a constant sum of surface areas to relate variables, then minimizing the sum of volumes.

Let the sphere have radius $r$.

Surface area of the parallelopiped:

$2(x\cdot 2x + 2x\cdot \frac{x}{3} + x\cdot \frac{x}{3}) = 2\left(2x^2 + \frac{2x^2}{3} + \frac{x^2}{3}\right) = 2\left(3x^2\right) = 6x^2$.

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

Given constant $k$:

Sum of volumes:

Substitute $r = \sqrt{\frac{k - 6x^2}{4\pi}}$ and differentiate $V$ w.r.t. $x$, set $\frac{dV}{dx}=0$.

After simplification (using $r$ to eliminate $k$), the condition reduces to $x = 3r$. …

This is a classic optimization problem where two shapes share a fixed total surface area, and we want to minimise their combined volume. The key is to use the constraint to eliminate one variable, leaving a single-variable function to minimise.

1. Write the given data and the constraint

The rectangular parallelepiped has sides $x$, $2x$, and $\frac{x}{3}$.

Its surface area is:

Let the sphere have radius $r$. Its surface area is $S_{\text{sphere}} = 4\pi r^2$.

The total surface area is constant; call it $k$:

2. Express $r$ in terms of $x$

From the constraint:

Since $r > 0$, we need $k > 6x^2$, which will hold for the relevant domain.

3. Write the sum of volumes

Volume of the box:

Volume of the sphere:

So the total volume is:

But $r$ is not independent — substitute $r = \left( \frac{k - 6x^2}{4\pi} \right)^{1/2}$:

Simplify the second term:

Thus:

4. Differentiate and set to zero

Differentiate with respect to $x$:

Simplify:

Set $V'(x) = 0$:

Since $x > 0$, divide by $x$:

Square both sides:

Multiply through by $\pi$:

Bring terms together:

So:

5. Show that $x = 3r$ at this critical point

From the constraint $6x^2 + 4\pi r^2 = k$, substitute $k$:

But from the critical condition we also have $k = \frac{(4\pi + 54)x^2}{9}$. Equate:

Multiply by 9:

Cancel $54x^2$ from both sides:

Divide by $4\pi$:

This proves the required relation.

6. Verify it’s a minimum

Take the second derivative:

At $x = 3r$, using $k - 6x^2 = 4\pi r^2$, we get $(k - 6x^2)^{1/2} = 2r\sqrt{\pi}$. Substituting and simplifying (the algebra is straightforward but lengthy) yields $V''(x) > 0$, confirming a minimum. …

Method: Optimizing a Combined Quantity for Two Shapes Sharing One Constraint

Some problems give you two separate shapes (here, a box and a sphere) whose individual surface areas or volumes are unrelated, but a single combined quantity (their total surface area, say) is held fixed. You're then asked to optimize a different combined quantity (their total volume). The technique is the same optimization skeleton, applied with two shape-formulas at once.

Steps

Step 1: Write each shape's surface area and volume in terms of its own defining variable.

Express everything the problem depends on (side length $x$ for the box, radius $r$ for the sphere) using the standard formulas for that shape.

Step 2: Write the shared constraint as a single equation equal to a constant.

Step 3: Write the objective — the combined quantity to optimize — as a function of both variables.

Step 4: Reduce to one variable, either by direct substitution or by Lagrange multipliers. …

Common Mistakes

Mistake 1: Miscounting the parallelopiped's surface area

Why it's wrong: With sides $x$, $2x$, $\dfrac{x}{3}$, the surface area is $2\left(x\cdot 2x+2x\cdot\dfrac{x}{3}+\dfrac{x}{3}\cdot x\right)=6x^2$ — students often forget the factor of $2$ (each pair of opposite faces counted once, then doubled) or miscompute one of the three face-pair products. Correct approach: list all three distinct face-pair areas first, sum them, then double the sum.

Mistake 2: Losing track of $k$ as a constant, not a value to solve for

Why it's wrong: $k=6x^2+4\pi r^2$ is given to be constant but its numeric value is never stated — the final minimum volume must stay expressed in terms of $k$ (or equivalently $r$). Treating $k$ as an unknown to be solved for, or dropping it partway through, produces a numerically meaningless "answer." Correct approach: carry $k$ symbolically throughout, and only substitute $r$'s relation to $k$ at the very end.

Mistake 3: Sign/chain-rule slip differentiating $r$ implicitly with respect to $x$ …