height = y = x^2
width = 2-x
area = a = x^2(2-x)
a = 2 x^2 -x^3
da/dx = 0 for max/min = 4 x -3x^2
0 = x (4 - 3 x)
x = 4/3
a rectangle with one side on the x axis and one side on the line x=2 has it upper left vertex on the graph of y=x^2, for what values of x does the area of the rectangle attain its maximum value?
1 answer