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UPSC 2022 Maths Optional Paper 1 Q2b — Step-by-Step Solution

15 marks · Section A

Lagrange's method of multipliers (constrained extrema) · Calculus · asked 8× in 13 yrs · Read the full method →

Question

A wire of length $l$ is cut into two parts which are bent in the form of a square and a circle respectively. Using Lagrange’s method of undetermined multipliers, find the least value of the sum of the areas so formed.

Technique

Standard Lagrange multipliers; gradient ratio $x/y=4/\pi$ gives the optimal split.

Solution

Setup. Let the wire be cut into pieces of length $x$ and $y$, with $x+y=l$.

• Square from piece $x$: side $x/4$, area $A_1=x^2/16$.
• Circle from piece $y$: circumference $y=2\pi r\Rightarrow r=y/(2\pi)$, area $A_2=\pi r^2=y^2/(4\pi)$.

Minimise $f(x,y)=A_1+A_2=\dfrac{x^2}{16}+\dfrac{y^2}{4\pi}$ subject to $g(x,y)=x+y-l=0$.

Step 1 — Set up Lagrangian

$L=f-\lambda g=\dfrac{x^2}{16}+\dfrac{y^2}{4\pi}-\lambda(x+y-l)$.

Stationary conditions:

• $\partial L/\partial x=x/8-\lambda=0\Rightarrow\lambda=x/8$.
• $\partial L/\partial y=y/(2\pi)-\lambda=0\Rightarrow\lambda=y/(2\pi)$.

Step 2 — Combine

$\dfrac{x}{8}=\dfrac{y}{2\pi}\Rightarrow y=\dfrac{2\pi x}{8}=\dfrac{\pi x}{4}$.

Step 3 — Apply constraint $x+y=l$

$x+\dfrac{\pi x}{4}=l\Rightarrow x\!\left(1+\dfrac{\pi}{4}\right)=l\Rightarrow x=\dfrac{4l}{4+\pi}$.

Then $y=\dfrac{\pi x}{4}=\dfrac{\pi l}{4+\pi}$.

Step 4 — Compute minimum area

Sum:

Answer