text: a rectangle has its base on the x-axis and its upper 2 vertices on the parabola y=12-x^2. what is the larges? t area the rectangle can have, and what are its dimensions? i don't remember that part of calculus very well but i think that this is the right way of making it.  the area of the rectangle is 2xy. i multiplied by 2 because xy would only calculate the area of the positive side of x. since its corners won't go besides the parabola we can use that function  to get y in terms of x in the rectangle. the area of the rectangle will now be 2(x)(12-x^2) a = 2x(12-x^2) a = 24x-2x^3 the derivative of the area is  a' = 24 - 6x^2 the critical points for the area are 0 = 24 - 6x^2 6x^2 = 24 x^2 = 4 x = 2 so the biggest area would be a(2) = 24(2) - 2 (2^3) a(2) = 48 - 16 = 32