A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is k (meter), then ( 4/π + 1 )k is equal to _____________.

Solution

Maths sir · 4 steps

Write the total area in terms of k alone, set its derivative to zero, and the answer falls out.

1. Set up areas. The circle uses length k, so its radius is k/2π and its area is k²/4π. The square uses the remaining 36-k, so its side is (36-k)/4 and its area is (36-k)²/16.

   A(k) = (36-k)²/16 + k²/4π

2. Differentiate. A'(k) = -(36-k)/8 + k/2π = 0 gives k/2π = (36-k)/8. Multiply both sides by 8 to get 4k/π = 36 - k.
3. Check minimum. A''(k) = 1/8 + 1/2π > 0, so this value of k gives a minimum.
4. So. Rearranging, 4k/π + k = 36, that is (4/π + 1)k = 36.

The idea to keep. In cut-the-wire problems, let one piece be the variable and write the other as total minus it. Then the whole area depends on one variable, and a single derivative finds the minimum.

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