- Draw a sketch of the region bounded above by y 1 = 8 − 2x^2 and below by y 2 = x^2 − 4, and
inscribe a rectangle in this region as described in the question. If (x, y 1 ) and (x, y 2 ) are the
vertices of the rectangle in quadrants I and IV, respectively, then the area
A = 2x (y 1 − y 2 ) = 2x(12 − 3x^2 ), or A(x) = 24x − 6x^3.
Then A′ (x) = 24 − 18x^2 = 6(4 − 3x^2 ), which equals 0 when Check to verify that A ′′
(x) < 0 at this point. This assures that this value of x yields maximum area, which is given by
dmanu
(dmanu)
#1