### JEE Main 12 Jan 2019 Morning Question 12

##### Question: The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, $ y=12-x^{2} $ such that the rectangle lies inside the parabola, is [JEE Main Online Paper Held On 12-Jan-2019 Morning]

#### Options:

A) 36

B) $ 18\sqrt{3} $

C) $ 20\sqrt{2} $

D) 32

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

Let (a, 0) be any point on the x-axis, which is the vertex of the rectangle

So, the co-ordinates of the vertex of the rectangle lying on the parabola $ y=12-x^{2} $ is $ (a,12-a^{2}). $

$ \therefore $ Area of rectangle, $ f(a)=2a(12-a^{2}) $

$ \therefore $ $ f’(a)=2(12-3a^{2}) $ For maximum area $ f’(a)=0 $

$ \Rightarrow $ $ 2(12-3a^{2})=0 $

$ \Rightarrow $ $ a=\pm 2 $

$ \therefore $ Maximum area at a=2 is f(2)=32 sq. units