Optimization problem

[tandoorichicken]

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

[himanshu121]

Since one side is on x-axis then other side will be || x-axis(rectangle) let say y=a then u will x=+sqrt(12-b).

Therefore u have length say = 2sqrt(12-b) &
breadth = b.

Hence Area, A = b\sqrt{12-b}.

u will get b=8 and Length=8& breadth 8.

Alternate
Rectangle with max area is a square
u get b=2\sqrt{12-b}