A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 3 − x 2 . What are the dimensions of such a rectangle with the greatest possible area?

Question

Width =   

Height = 

Answer 1

The set up is to find a formula for the rectangle area as a function of x, the length of the base of the rectangle in the +x direction.

A = (2x)(2y) = 2x*2(3-x²) = 12x - 4x³ ( note that A is 0 for x = 0 or sqrt(3)

You could graph this and pick off the max with trace. Desmos highlights 0s and extrema. (It gives an integer value which you can check to be a max)

The method in calculus is to find dA/dx = 0 or 12 -12x² = 0 x = +/- 1 (take the positive value) and A_max = 8

You can show it's a max by the derivative going from + to - around the point x=1. You can also find that the second derivative is negative - implying a max.

Answer 2

vetex of the parabola is (0,3)

right half of the rectangle has base from (0,0) to (x,0)

height = y from (0,0) to (0,y) and from (x,0) to (x,y)

Area of the half rectangle = A = xy

y = 3-x^2

A= xy = x(3-x^2) = 3x - x^3

A' = 3-3x^2 = 0

x^2= 1 or -1

x=1

y = 3-(+/-1)^2 = 3-1 = 2

Area = xy = 1(2) = 2 = max area

for half the area under the parabola

multiply it by two to get

4 = maximum area

vertices of the full rectangle are (-1,2), (1,2), (-1,0), (1,0)

dimensions are width = 2, height = 2