A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola #y=6-x^2#. What are the dimensions of such a rectangle with the greatest possible area?

thanks for any help!?

Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola #y=6-x^2#. What are the dimensions of such a rectangle with the greatest possible area?

thanks for any help!?

Isaiah Allen · Accepted Answer

Height: #4# units
Width #2sqrt(2)# Start by sketching #y=6 -x^2#. Then draw a rectangle beneath it. You will notice that the width is #2x# and the height is #6-x^2#. Area is given by length times width, so the area function will be #A=2x(6-x^2) =12x - 2x^3#. Now you differentiate to find the maximum. #A’ =12 -6x^2# Find critical numbers by setting #A’ # to #0#. #x =+-sqrt(2)# The derivative is negative at #x=2# and positive at #x=1#, which justifies that the rectangle with width of #sqrt(2)# has maximal area. The height will be #y(sqrt(2)) = 6-(sqrt(2))^2) = 4# Hopefully this helps!

Emilia Aguilar · Answer

undefined The dimensions of the rectangle with the greatest possible area are: 
Width: 2√6 units 
Height: 12 units

Isaiah Allen · Answer

undefined To find the dimensions of the rectangle with the greatest possible area, we first need to find the coordinates of the two points where the rectangle touches the parabola.

1. The parabola equation is y = 6 - x^2.
2. Since the base of the rectangle is on the x-axis, the y-coordinate of the points where the rectangle touches the parabola will be 0.
3. Set y = 0 in the parabola equation: 0 = 6 - x^2.
4. Solve for x: x^2 = 6.
5. Take the square root of both sides: x = ±√6.

So, the x-coordinates of the two points where the rectangle touches the parabola are ±√6.

The width of the rectangle is the difference between these x-coordinates: 2√6.

To find the height of the rectangle, substitute one of the x-values (±√6) into the parabola equation:

When x = √6:
y = 6 - (√6)^2 = 6 - 6 = 0.

When x = -√6:
y = 6 - (-√6)^2 = 6 - 6 = 0.

Since the y-coordinate of both points is 0, the height of the rectangle is also 0.

Therefore, the dimensions of the rectangle with the greatest possible area are:
Width = 2√6
Height = 0