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. 

24 is the correct answer.