Arectangle is constructed with its base on the x-axis and two of its vertices on the parabola y=121-x^2. what are the dimensions of the rectangle with the maximum area?

The dimensions of the rectangle with maximum area is given by x and y

values at the maximum value of the function for its area.

The correct responses are;

Reasons:

The given function on which the vertices is located is y = 121 - x²

The coefficient of x in the given function is 0, therefore, the location of the

x-coordinate of the vertex of the parabola is on the y-axis.

The height of the rectangle, y = 121 - x²

The rectangle will extend equally on both sides of the y-axis, therefore;

The width of the rectangle = 2·x

The area of the rectangle, A = 2·x × (121 - x²) = 242·x - 2·x³

At the maximum area, we have;

Which gives;

The height of the rectangle, y = 121 - x²

The height of the rectangle, y =

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the length is and the width is

Step-by-step explanation:

Let points A and B be placed on the x-axis. Their coordinates are and (because of parabola symmetry). Two other vertices lie on the parabola, then and The length of the side AB is and the length of the side AD is Thus, the area of the rectangle ABCD is

Find the derivative A':

Equate A' to 0:

The maximum area of the rectangle is

The dimensions of the rectangle are:

the length is and the width is

the figures below are made out of circles, semicircles, quarter … find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations). remember that we'll also have to cut whatever areas we get in half, since we're only dealing im pretty sure that is the right answer

step-by-step explanation:

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