Find the area of the largest rectangle (with sides parallel to the coordinate axes) that can be inscribed in the region enclosed by the graphs of f (x) = 18-x^2 and

F (x) = 18-x^2 is a parabola having vertex at (0, 18) and opening downwards.

g (x) = 2x^2-9 is a parabola having vertex at (0, - 9) and opening upwards.

By symmetry, let the x-coordinates of the vertices of rectangle be x and - x = > its width is 2x.

Height of the rectangle is y1 + y2, where y1 is the y-coordinate of the vertex on the parabola f and y2 is that of g.

=> Area, A

= 2x (y1 - y2)

= 2x (18 - x^2 - 2x^2 + 9)

= 2x (27 - 3x^2)

= 54x - 6x^3

For area to be maximum, dA/dx = 0 and d²A/dx² < 0

=> 54 - 18x^2 = 0

=> x = √3 (note: x = - √3 gives the x-coordinate of vertex in second and third quadrants)

d²A/dx² = - 36x < 0 for x = √3

=> maximum area

= 54 (√3) - 6 (√3) ^3

= 54√3 - 18√3

= 36√3.