A rectangle is placed under the parabolic arch given by f(x) = 27 - 3x2 by using a point (x, y) on the parabol

Question

A rectangle is placed under the parabolic arch given by f(x) = 27 - 3x2 by using a point (x, y) on the parabola, as shown in the figure. Write a formula for the function A(x) that gives the area of the rectangle as a function of the x-coordinate of the point chosen.

Answer 1

The y-value for any given x-value is 27 - 3x^2. That's supplied up-front. A rectangle from the origin to any point (x,y) has area x*y.

So... the portion of the inscribed rectangle on the positive-x side of the y-axis has area x*y, or x(27 - 3x^2).

There's a matching rectangle on the negative-x side of the y-axis with the same area.

This makes the overall area two times the x-value times the y-value of the point on the positive-x side of the y-axis:

2 * x * (27 - 3x^2)

Which is:

A(x) = 54x - 6x^3

Answer 2

You say under a parabolic arch but do not say where the arch ends. At the x-axis? At the line y=-4? At infinity?

This parabola has a maximum of 27 located at (0,27) which is also the vertex. The axis of symmetry is the y-axis. The x-intercepts are (-3,0) and (3,0). The range is y =< 27.

Unless we know where the arch ends, al we can say is A = infinity.

Answer 3

y = 27 - 3 x^2 . . . parabola is with vertex (0,27) . . . facing down
assuming the lower boundary is the x-axis

area = 2 x y = 2 x [ 27 - 3 x^2 ] = 54 x - 6 x^3