A rectangle is inscribed with its base on the x -axis and its upper corners on the parabola
y= 11-x^2.
What are the dimensions of such a rectangle with the greatest possible area?
The area of the rectangle with corners (-x,0), (x,0) (-x,y) (x,y) is
a = 2x * (11-x^2)
a = 22x - 2x^3
da/dx = 22 - 6x^2
da/dx = 0 when x = √(11/3)
So, the rectangle is 2√(11/3) x 22/3 with area 44/3 * √(11/3) = 28.08
To find the dimensions of the rectangle with the greatest possible area, we need to find the coordinates of the points where the rectangle's upper corners touch the parabola.
Let's assume the base of the rectangle on the x-axis has a length of 2a. This means the rectangle's upper corners have coordinates (a, 11 - a^2) and (-a, 11 - a^2).
The area of the rectangle is given by the length of the base multiplied by the height. In this case, the area is A = 2a * (11 - a^2).
To find the dimensions of the rectangle with the greatest possible area, we need to find the value of a that maximizes the area. We can do this by taking the derivative of the area function with respect to a, setting it equal to zero, and solving for a.
Let's differentiate the area function with respect to a:
dA/da = 2(11 - a^2) - 2a(2a) = 22 - 4a^2 - 4a^2 = 22 - 8a^2 - 4a^2
Now, let's set dA/da equal to zero and solve for a:
22 - 8a^2 - 4a^2 = 0
Combining like terms:
-12a^2 + 22 = 0
Rearranging the equation:
12a^2 = 22
a^2 = 22/12
a^2 = 11/6
Taking the square root of both sides:
a = ± √(11/6)
Since we are dealing with a rectangle, the length cannot be negative. Therefore, we take a = √(11/6).
Now we can plug this value of a back into the area function to find the maximum area:
A = 2a * (11 - a^2)
A = 2√(11/6) * (11 - (11/6))
A = 2√(11/6) * (66/6 - 11/6)
A = 2√(11/6) * (55/6)
A = (110/6)√(11/6)
A ≈ 20.30
So, the rectangle with the greatest possible area has dimensions approximately 2√(11/6) by 110/6.