Answer to Question #193460 in Calculus for Ok099

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