Arectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=6−x2. what are the dimensions of such a rectangle with the greatest possible area?

length: 2sqrt(2)

width: 4

Step-by-step explanation:

Let the x-coordinate of the right lower vertex be x. Then the x-coordinate of the left lower vertex is -x. The distance between x and -x is 2x, so the lower side of the rectangle, that rests on the x-axis, measures 2x. The length of the rectangle is 2x.

We now use the equation of the parabola to find the y-coordinate of the upper vertices. y = 6 - 2^x. If you plug in x for x, naturally you get y = 6 - x^2, so the y-coordinates of the two upper vertices are 6 - x^2. Since the y-coordinates of the lower vertices is 0, the vertical sides of the rectangle have length 6 - x^2. The width of the rectangle is 6 - x^2.

We have a rectangle with length 2x and width 6 - x^2.

Now we can find the area of the rectangle.

area = length * width

A = 2x(6 - x^2)

A = 12x - 2x^3

To find a maximum value of x, we differentiate the expression for the area with respect to x, set the derivative equal to zero, and solve for x.

dA/dx = 12 - 6x^2

We set the derivative equal to zero and solve for x.

12 - 6x^2 = 0

2 - x^2 = 0

x^2 = 2

x = +/- sqrt(2)

length = 2x = 2 * sqrt(2) = 2sqrt(2)

width = 6 - x^2 = 6 - (sqrt(2))^2 = 6 - 2 = 4

where is abdul's work? don't see that anywhere posted : /

step-by-step explanation: