Calculus Homework- Dimensions of a rectangle with the greatest possible area.

Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=10−x^2. What are the dimensions of such a rectangle with the greatest possible area?


Width: ___

Height: 20/3

 

I can't figure out the width, the height is right. I've tried 10/3, (1/3)^30, (2/3)^30 None of those answers have been right.

Answer 1

The coordinates of the upper right corner of the rectangle are (x, 10-x²).

The dimensions of the rectangle are 2x by 10-x^2.

 

A(x) = area of rectangle

 

       = 2x(10-x²) = -2x³ + 20x

 

A'(x) = -6x²+20

 

A'(x) = 0 when x = √(10/3)

 

To the left of x = √(10/3), A'(x) > 0 and to the right of x = √(10/3), A'(x) < 0.  So, A(x) is maximized when x = √(10/3).

 

Dimensions of rectangle:  2x = 2√(10/3)

 

                                    10 - x² = 10 - 10/3 = 20/3