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?
Width=? Height=?
Area= height*width=y(2x)
Area=(11-x^2)(2x)
solve for dArea/dx, set to zero, solve for x. Width will be 2x, height y.
ty very much!
You're welcome! To find the dimensions of the rectangle with the greatest possible area, we need to maximize the area function. The area of a rectangle is given by the formula A = height * width.
In this case, since the base of the rectangle is on the x-axis and the upper corners are on the parabola y = 11 - x^2, we can say that the height of the rectangle is y and the width is 2x (twice the value of x because the corners are symmetric about the y-axis).
So, the area function can be written as A = y * (2x).
Substituting the equation of the parabola into the area function, we get A = (11 - x^2) * (2x).
To find the maximum area, we need to calculate the derivative of the area function with respect to x, set it equal to zero, and solve for x.
Taking the derivative of A with respect to x:
dA/dx = d/dx [(11 - x^2) * (2x)]
= 2(11 - x^2) * (1) + (11 - x^2) * (2)
= 22 - 2x^2 + 22x - 2x^3
Setting dA/dx equal to zero:
22 - 2x^2 + 22x - 2x^3 = 0
Now, we solve this equation to find the value(s) of x. After finding the value(s) of x, we can substitute it back into the area function to find the width (2x) and height (y) of the rectangle.
By solving the equation for x, we can find the x-coordinate(s) of the points at which the rectangle's area is maximized.