Since this is an upside down parabola, the rectangle must be inside the parabola. The parabola and rectangle inside it are symmetrical about the y axis.
Let x be the coordinate of a corner that touches the parabola at y = 12 - x^2. The area of that parabola is
A = 2x*(12-x^2)= 24 x - 2x^3
That is a maximum when dA/dx = 0
24 -6x^2 = 0
x = +/- 2
y = 12 - 4 = 8
Maximum area = 2xy = 32
A rectangle has its base on the x-axis and its 2 upper corners on the parabola y=12-x^2. What is the largest possible area of the rectangle?
1 answer
1 answer