Find the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve `y=12-x^2.`

Options

A. 36
B. `20sqrt(2)`
C. 32
D. `18sqrt(3)`

Answer

C

Explanation

Equation of parabola is given, `y=12-x^(2)`
or `x^(2)=-(y-12)`
Not that vertex of parabola is (0,12) and its open donward.
Let Q be one of the vertices of rectangle which lies on parabola. Then, the coordinates of Q be (a,12-a^(2))`
ltrbgtThen area of rectangle PQRS
`=2xx` (Area of rectangle PQMO)
[ due to symmetry about Y-axis]
`=2xx[a(12-a^(2))]=24a -2a^(3)=Delta` (let).
The area function `Delta_(a)` will be maximum, when
`(d Delta)/(da)=0`
`rArr 24-6a^(2)=0`
`rArr a^(2)=4rArr a=2 " " [ :. agt 0]`
So, maximum area of rectangle
`PQRS=(24x x2)-2(2)^(3)`
`=48-13=32` sq units