JEE Main 2021 · Mathematics · Differentiation and Application of Derivatives · medium

A rectangle has two vertices on the x-axis and the other two on the semicircle y=√(25-x^2. The maximum possible area of the rectangle is

Single correct option

A rectangle has two vertices on the

x

-axis and the other two on the semicircle

y=\sqrt{25-x^2}

. The maximum possible area of the rectangle is

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Answer: (B) $25$

By symmetry take the vertices on the semicircle as

(\pm x,\sqrt{25-x^2})

with

0<x<5

. The area is

A(x)=2x\sqrt{25-x^2}

. Maximise

A^2=4x^2(25-x^2)

: setting its derivative

4(50x-4x^3)

to zero gives

x^2=\dfrac{25}{2}

. Then

\sqrt{25-x^2}=\dfrac{5}{\sqrt2}

and

x=\dfrac{5}{\sqrt2}

, so

A=2\cdot\dfrac{5}{\sqrt2}\cdot\dfrac5{\sqrt2}=25

. This is a maximum because

A\to0

at both ends

x\to0,5

. The square-like case

x=3

gives only

24

Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake