(1 point) A rectangle is inscribed with its base on the xx-axis and its upper corners on the parabola y=11−x2y=11−x2. What are the dimensions of such a rectangle with the greatest possible area?

the rectangle has dimensions of length= 2√(11/3) and height=22/3

Step-by-step explanation:

since the parabola

y= 11-x²

is symmetric with respect to the y-axis ( meaning that y(x)=y(-x) );

area of rectangle with base from -x to x and height y = 2 *area of rectangle with base from 0 to x and height y

A=2*x*y

replacing y

A=2*x*( 11-x²) = 22*x - 2*x³

the maximum value can be found when the derivative of the area with respect to x is 0 , thus

dA/dx= 22 - 6*x² = 0

22 - 6*x² = 0

x max = √(22/6)

y max = 11-xmax² =  11-[√(22/6)]²= 11 - 22/6 = 44/6

then the rectangle with maximum area has dimensions of

length = 2*x max= 2√(22/6) and height=y max=44/6