A wire of length 20 m is to be cut into two pieces. A piece of length l₁ is bent to make a square of area A₁ and the other piece of length l₂ is made into a circle of area A₂. If 2 A₁+3 A₂ is minimum then (π l₁): l₂ is equal to :

Solution

Maths sir · 4 steps

Write both areas in terms of the wire lengths, use l₂ = 20 - l₁, then differentiate once.

1. Areas from lengths. The square has side l₁/4, so A₁ = l₁²/16. The circle has 2π r = l₂, so r = l₂/2π and A₂ = π r² = l₂²/4π.

   f = 2A₁ + 3A₂ = l₁²/8 + (3 (20 - l₁)²)/4π

2. Set derivative zero. df/dl₁ = l₁/4 - 3(20 - l₁)/2π = l₁/4 - 3l₂/2π = 0. Multiply through by 4π to get π l₁ = 6 l₂.
3. Check it is minimum. (d² f)/dl₁² = 1/4 + 3/2π > 0, so this point gives a minimum.
4. So. π l₁ : l₂ = 6 : 1, which is option (a).

The idea to keep. In wire-cutting problems, turn every area into a function of the piece lengths, use the fixed total to keep one variable, and set the derivative to zero. Carry any weights on the areas all the way through.

Answer from the NTA final answer key. Solution by PrepWiser.