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

Answer 1

Make a sketch and draw in the rectangle.
Let the top-right point be P(x,26-x^2)
and the bottom-right point be Q(x, x^2 + 2)

So the width of the rectangle is 2x
and the height is 26-x^2 - x^2 - 2
= 24 - 2x^2

area = 2x(24-2x^2)
= 48x - 4x^3
d(area)/dx = 48 - 12x^2
= 0 for a max of area
12x^2 = 48
x^2 = 4
x = 2

the largest area = 48(2) - 4(2^3)
= 96 - 32
= 64 units^2

Answer 2

thank you so much!