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In the revenue equation R = –360p2 + 28,800p, what is the maximum revenue that can be expected?

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In the revenue equation R = –360p2 + 28,800p, what is the maximum revenue that can-example-1

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The revenue equation is given by:

R = –360p^2 + 28,800p

To find the maximum revenue that can be expected, we need to find the vertex of the parabola represented by this equation. The vertex of a parabola is given by the formula:

Vertex = (-b/2a, f(-b/2a))

where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.

In this case, a = -360 and b = 28,800.

The x-coordinate of the vertex is given by:

x = -b/2a = -28,800/(2*(-360)) = 40

Substituting x = 40 into the revenue equation gives:

R = –360(40)^2 + 28,800(40) = 576,000

Therefore, the maximum revenue that can be expected is $576,000.

Again let me know what other ones you need help with

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