A rectangle is bounded by the x-axis and the semicircle y = radical 9 − x^2 (see figure). What length and width should the rectangle have so that its area is a maximum?

Answer 1

Area = A = xy

A = xsqr(9-x^2)

take the derivative and set = 0

A' = (x/sqr(9-x^2)(-2x) + sqr(9-x^2) = 0

multiply by sqr(9-x^2) to get

A' = -x^2 + 9 -x^2 = 0

-2x^2 =-9, x^2 = 9/2, x = 3/sqr2 = (3/2)sqr2 = about 2.121, 2x = 4.242

y = sqr(9-x^2) = sqr(9-9/2) = 3sqr2/2 = 2.121

dimensions of the area maximizing rectangle is 4.242 by 2.121 or exactly 3sqr2 by (3/2)sqr2

max Area = exactly 9