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=?
To find the dimensions of the rectangle with the greatest possible area, we need to maximize the area of the rectangle. The area of a rectangle is given by the formula: Area = Length × Width.
In this case, the base of the rectangle is on the x-axis, so the length of the rectangle will be the x-coordinate of the upper corners of the rectangle. Let's denote this x-coordinate as "x".
We are also given the equation of the parabola, y = 11 – (x^2). Since the upper corners of the rectangle lie on this parabola, the height of the rectangle will be the y-coordinate, which is given by the equation of the parabola.
Now, we can express the area of the rectangle as a function of x:
Area = Length × Width = x × (11 – (x^2))
To find the maximum area of the rectangle, we need to find the value of x that maximizes this function.
To find the maximum of a function, we can take its derivative and set it equal to zero.
Let's differentiate the function with respect to x:
d(Area)/dx = d(x(11 - (x^2)))/dx
= 11 - 3x^2
Setting the derivative equal to zero:
11 - 3x^2 = 0
Solving this equation:
3x^2 = 11
x^2 = 11/3
x = ± sqrt(11/3)
Since we are looking for a rectangle on the x-axis, the negative value of x is not applicable in this case. Hence, we take the positive value of x.
So, the width of the rectangle is given by x, and the height of the rectangle is given by the y-coordinate on the parabola, which is:
y = 11 – (x^2)
= 11 – (11/3)
= 22/3
Therefore, the dimensions of the rectangle with the greatest possible area are:
Width = x = sqrt(11/3)
Height = y = 22/3