Q.A metal box with a square base and vertical sides is to contain $1024$ cm$^3$. The material for the top and bottom costs Rs $5$/cm$^2$ and the material for the sides costs Rs $2.50$/cm$^2$. Find the least cost of the box.

Concept: Optimization Word Problem — minimize cost given a fixed volume constraint.

Let the side of the square base be $x$ cm and the height be $h$ cm.

Volume: $x^2 h = 1024 \Rightarrow h = \frac{1024}{x^2}$.

Cost function:

Top + bottom area = $2x^2$, cost at Rs 5/cm² → $10x^2$.

Four sides area = $4xh$, cost at Rs 2.50/cm² → $10xh$.

So total cost $C = 10x^2 + 10x \cdot \frac{1024}{x^2} = 10x^2 + \frac{10240}{x}$.

Differentiate: $C'(x) = 20x - \frac{10240}{x^2}$. Set $C'(x)=0$: …

We have a box with a square base. Let the side of the square base be $x$ cm and the height be $h$ cm. The volume is fixed at $1024$ cm$^3$, so:

The cost has two parts: top and bottom (area $2x^2$ at Rs $5$/cm$^2$) and the four sides (area $4xh$ at Rs $2.50$/cm$^2$). So the total cost $C$ in rupees is:

Substitute $h$:

We need to minimize $C(x)$ for $x > 0$.

1. Find the derivative Differentiate $C(x)$ with respect to $x$:

2. Set derivative to zero

Multiply both sides by $x^2$:

3. Verify it's a minimum The second derivative is:

At $x = 8$, $C''(8) = 20 + \frac{20480}{512} = 20 + 40 = 60 > 0$, so it's a local minimum. Since $C(x) \to \infty$ as $x \to 0^+$ and as $x \to \infty$, this is the global minimum.

4. Find the height and cost $h = \frac{1024}{8^2} = \frac{1024}{64} = 16 \text{ cm}$ …

Method: Minimizing Total Cost Under a Fixed-Volume Constraint

This technique applies to any "container" problem where the volume is fixed and you must minimize the material cost (or surface area) — the classic box/can/tank family of optimization problems.

Steps

Step 1: Name the container's dimensions and write the volume constraint.

Assign a variable to each independent dimension (for a box with a square base, one side length $x$ and a height $h$ are enough). Write the fixed volume as an equation in these variables, e.g.

Step 2: Solve the constraint for one variable, so only one is left free.

Make the "harder to eliminate" variable (usually the height, which appears linearly) the subject:

Step 3: Write the total cost (or surface area) as a function of the remaining variable.

Identify every distinct surface (top+bottom, the four sides, etc.), multiply each area by its own per-unit rate, add them, then substitute Step 2's expression so the cost depends on a single variable:

Step 4: Differentiate and solve $C'(x)=0$. …

Common Mistakes

Mistake 1: Forgetting the box has TWO faces (top and bottom), not one

Why it's wrong: The top and bottom are each a square of area $x^2$, so their combined material costs $5(2x^2)=10x^2$, not $5x^2$. A student who only counts one square face undercounts the top/bottom cost by half. Correct approach: always count all six faces explicitly — 2 faces of area $x^2$ and 4 faces of area $xh$ — before multiplying by the rate.

Mistake 2: Forgetting the box has FOUR side faces, not one

Why it's wrong: The four vertical sides each have area $xh$, so their total cost is $2.50(4xh)=10xh$, not $2.50(xh)$. Missing the factor of 4 gives a cost function that is off by a large constant multiple and leads to the wrong critical value of $x$. Correct approach: write total side area as $4xh$ before applying the per-$\text{cm}^2$ rate.

Mistake 3: Solving $C'(x)=0$ and stopping, without confirming it's a minimum …