Expert: Sherry Wallin - 5/5/2009

Question
5.)Find the dimensions of the rectangle with the largest area inscribed inside the area bounded by y = 16 - x^2 and y = 0 if one side of the rectangle must coincide with x - axis

Answer
Hi Raf~
    The area of a rectangle is 2xy. Eliminate y from 2xy by substituting y = 16-x^2 into 2xy. So 2xy = 2x(16-x^2) = 32x - 2x^3. Take the first derivative of f(x) = 32x - 2x^3 so f'(x) = 32-6x^2. Now factor it to find the critical points: 2(16 - 3x^2) = 0 -> 3x^2 = 16 ->
x^2 = 16/3 -> x = +-[4/sqrt(3)]= +-[4*sqrt(3)/3]. Just remember you are looking for the maximum so you want to know what interval is positive. The intervals you get are (-inf, -4*sqrt(3)/3],[-4*sqrt(3)/3, 4*sqrt(3)/3], and [4*sqrt(3)/3, inf) and the only interval that is positive is [-4*sqrt(3)/3, 4*sqrt(3)/3] which is where your maximum is. Now plug x = 4*sqrt(3)/3 into y = 16-x^2 and find that y = 32/3 so the dimensions of the rectangle with the largest area inscribed inside the area bounded by y = 16-x^2 and y = 0 is 4*sqrt(3)/3 units by 32/3 units.

Math Prof