Q85 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics

A wire 40 metre in length is to be cut into two pieces. One piece is formed into a square and the other piece into a circle. The lengths of the two pieces so that the combined area of the square and the circle is minimum, are respectively

  • Correct answer
    A
    $\dfrac{160}{\pi + 4}, \dfrac{40\pi}{\pi + 4}$
  • B
    $\dfrac{\pi + 4}{160}, \dfrac{40\pi}{\pi + 4}$
  • C
    $\dfrac{160}{\pi + 4}, \dfrac{\pi + 4}{40\pi}$
  • D
    $30, 10$

Solution

Let the length of the piece for the square be $x$ and for the circle be $40 - x$.

Perimeter of the square $= x \Rightarrow$ side $= \dfrac{x}{4}$

Area of the square, $A_1 = \dfrac{x^2}{16}$

Circumference of the circle $= 40 - x \Rightarrow$ radius $r = \dfrac{40 - x}{2\pi}$

Area of the circle, $A_2 = \pi r^2 = \dfrac{(40 - x)^2}{4\pi}$

Total area, $A = \dfrac{x^2}{16} + \dfrac{(40 - x)^2}{4\pi}$

For minimum area, $\dfrac{dA}{dx} = 0$

$\dfrac{2x}{16} - \dfrac{2(40 - x)}{4\pi} = 0$

$\dfrac{x}{8} = \dfrac{40 - x}{2\pi}$

$\pi x = 160 - 4x$

$x(\pi + 4) = 160 \Rightarrow x = \dfrac{160}{\pi + 4}$

Length of the other piece $= 40 - x = 40 - \dfrac{160}{\pi + 4} = \dfrac{40\pi}{\pi + 4}$

Since $\dfrac{d^2A}{dx^2} = \dfrac{1}{8} + \dfrac{1}{2\pi} > 0$, the area is minimum for these lengths.

Answer: $\dfrac{160}{\pi + 4}, \dfrac{40\pi}{\pi + 4}$

Hence, the correct answer is option: 'A'.

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