Q.An open box with a square base is to be made out of a given quantity of cardboard of area $c^2$. Show that the maximum volume of the box is $\dfrac{c^3}{6\sqrt{3}}$ cubic units.

Concept: Optimization Word Problem — maximizing volume under a fixed surface area constraint.

Let the square base have side $x$ and the height be $h$. The cardboard area (open box) is:

So $h = \dfrac{c^2 - x^2}{4x}$. Volume $V = x^2 h = \dfrac{x(c^2 - x^2)}{4}$.

Differentiate $V$ with respect to $x$ and set to zero:

$\frac{dV}{dx} = \frac{c^2 - 3x^2}{4} = 0 \quad\Rightarrow\quad x^2 = \frac{c^2}{3}$ …

This is a classic optimization problem: you have a fixed amount of material (cardboard of area $c^2$) and you want to shape it into an open-top box (square base, no lid) that holds the greatest possible volume. The key is to translate the physical constraint — the given area — into an equation linking the dimensions, then write volume as a function of a single variable and find its maximum using differentiation.

Let’s go step by step.

1. Define the variables.

Let the side of the square base be $x$ units, and let the height of the box be $h$ units. Since the box is open at the top, the cardboard used consists of:

• the base: area $x^2$
• four side walls: each of area $x h$, so total side area $4xh$

The total area of cardboard is given as $c^2$. So:

2. Express height in terms of base side.

From the area constraint:

We need $x > 0$ and $h > 0$, so $c^2 - x^2 > 0$, i.e. $0 < x < c$.

3. Write the volume function.

Volume $V$ of the box is base area times height:

So:

4. Maximize $V(x)$ for $0 < x < c$.

Differentiate with respect to $x$:

Set $V'(x) = 0$:

(The negative root is irrelevant.)

5. Confirm it’s a maximum.

Check the second derivative: …

Method: Maximizing the Volume of a Container Built from a Fixed Amount of Material

This is the standard box-from-a-sheet optimization family: express the material used as a constraint equation, use it to eliminate one dimension, then maximize the resulting single-variable volume function.

Steps

Step 1: Name the dimensions and write the material (area) constraint

Assign a variable to each independent dimension of the container (e.g. base side, height). Add up the areas of every actual piece of material used — base, walls, and a lid if present — and set that sum equal to the given fixed area.

Step 2: Solve the constraint for one dimension and substitute into the volume formula

Solve the area equation for the less convenient variable (typically height, since it usually appears linearly) in terms of the other. Substitute this into the volume formula so that volume becomes a function of a single variable.

Step 3: Differentiate, solve, and confirm it is a maximum …

Common Mistakes

Mistake 1: Writing the area equation as if the box had a lid

Since the box is open (no top), the cardboard used is only the base plus four sides: $x^2+4xh=c^2$. Why it's wrong: forgetting "open" and including a top face gives $2x^2+4xh=c^2$ instead, which changes every downstream step and produces a wrong maximum volume. Correct approach: re-check whether a box problem says "open" or "closed" and count faces accordingly — one square base, four rectangular sides, zero tops, for this problem.

Mistake 2: Losing track of the domain $0<x<c$ and accepting a negative or invalid root …