Question

Find the area of the largest rectangle with sides parallel to the coordinate axes which can be inscribed in the bounded by the two parabolas y=26-x^2 and y=x^2+2

Answer 1

The graph of the above function is-

Each side intersect both functions twice

so area of rectangle = length * width

length = $(26-x^2) - (x^2+2) = 24 - 2x^2$

width = $-2x$

area of rectangle $= (24-2x^2)\times -2x$

$A=4x^3-48x$

taking the first derivative of the area

$A'=12x^2 - 48$

Putting $A'=0, 12x^2=48\Rightarrow x^2=4\Rightarrow x=\pm 2$

Again differentiating A'=

$A''=24x \\ \text{ at } x=-2, A''=-48<0$

So Area is maximum at x=-2

Maximum area =$4(-2)^3-48(-2)=-32+96=64$