Q.A wire of length $28$ m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Concept: Lagrange Multipliers (or single-variable optimisation).

Let the two pieces be $x$ m (square) and $28-x$ m (circle).

Side of square $= \frac{x}{4}$, area $= \frac{x^2}{16}$.

Radius of circle $= \frac{28-x}{2\pi}$, area $= \frac{(28-x)^2}{4\pi}$.

Total area $A(x) = \frac{x^2}{16} + \frac{(28-x)^2}{4\pi}$.

Differentiate and set to zero:

$A'(x) = \frac{x}{8} - \frac{28-x}{2\pi} = 0$

$\Rightarrow \frac{x}{8} = \frac{28-x}{2\pi} \Rightarrow \pi x = 4(28-x)$

$\Rightarrow \pi x = 112 - 4x \Rightarrow x(\pi+4) = 112$ …

Why Lagrange multipliers? The big idea

You have a fixed resource — 28 metres of wire — and you must split it into two parts. Each part creates a shape with an area that depends on its perimeter. The total area is what you want to minimise. This is a classic constrained optimisation problem: you have one variable (the cut point) and one constraint (the total length is fixed). Lagrange multipliers are the elegant way to handle this when the constraint is an equation linking the variables.

But here, because there's only one free variable, you can also just substitute the constraint directly and use ordinary calculus. Both methods give the same answer. I'll show you the substitution method first — it's simpler — then mention how Lagrange multipliers would look.

Step-by-step solution

1. Define the variables

Let the piece used for the square have length $x$ metres. Then the remaining piece for the circle has length $28 - x$ metres.

Clearly, $0 \le x \le 28$, but the optimum will lie strictly inside this interval.

2. Express the areas in terms of $x$

• Square: Perimeter $= x$. A square has 4 equal sides, so each side $= \frac{x}{4}$.

  Area of square: $A_s = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}$.

• Circle: Perimeter (circumference) $= 28 - x$.

  Circumference $= 2\pi r$, so $r = \frac{28 - x}{2\pi}$.

  Area of circle: $A_c = \pi r^2 = \pi \left(\frac{28 - x}{2\pi}\right)^2 = \frac{(28 - x)^2}{4\pi}$.

3. Write the total area function

This is the function we need to minimise for $x \in [0, 28]$.

4. Differentiate and set to zero

Set $A'(x) = 0$:

Cross-multiply:

5. Check that this is a minimum

The second derivative:

$A''(x) = \frac{1}{8} + \frac{1}{2\pi} > 0$ …

Method: Splitting a Fixed Length Between Two Shapes

This method applies to any problem where a single fixed length (a wire, a rope, a budget) is divided between two different shapes, and you must minimize (or maximize) the combined quantity — here, area.

Steps

Step 1: Name one piece with a variable; the other follows from the total

If the total length is $L$ and one piece (say, used for the square) has length $x$, the other piece (for the circle) automatically has length $L - x$. You now have exactly one free variable, $x$.

Step 2: Convert each piece's length into that shape's own defining measurement

• A square's perimeter is $4\times(\text{side})$, so side $= \dfrac{x}{4}$, and its area is $\left(\dfrac{x}{4}\right)^2$.
• A circle's circumference is $2\pi r$, so radius $= \dfrac{L-x}{2\pi}$, and its area is $\pi\left(\dfrac{L-x}{2\pi}\right)^2$.

Do not confuse the piece length itself with the side or radius — always divide by the shape's own perimeter formula first.

Step 3: Write the total area as a single function of $x$ …

Common Mistakes

Mistake 1: Using the piece length as if it were the radius directly

The length $28 - x$ is the circle's circumference, not its radius. A student who writes the circle's area as $\pi(28-x)^2$ (skipping the $r = \dfrac{28-x}{2\pi}$ step) gets a completely wrong function to differentiate. Always convert perimeter/circumference to the shape's own linear measurement first.

Mistake 2: A sign slip differentiating $(28-x)^2$ …