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A company that sells digital cameras has found that revenue can be modeled by the equation R(p) = -5p^2 ¤ +1230p, WHERE P IS THE PRICE OF THE CAMERA IN DOLLARS? ALGEBRAICALLY DETERMINE WHAT PRICE WILL MAXIMUM THE REVENUE? WHAT IS THE MAXIMUM REVENUE?​

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Answer:

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Step-by-step explanation:

User Dmmfll
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Final Answer:

The company will maximize their revenue by setting the camera price at $61.50. At this price, the maximum revenue will be $37,987.50.

Step-by-step explanation:

Revenue function: R(p) = -5p^2 + 1230p. This is a quadratic function with a negative leading coefficient, indicating a downward facing parabola.

Maximum revenue: To find the price that maximizes revenue, we need to find the vertex of the parabola. The vertex occurs at the price where the derivative R'(p) = 0.

Differentiate: R'(p) = -10p + 1230.

Set derivative to zero and solve for p: -10p + 1230 = 0 --> p = 123.

Check if it's a maximum: Since the leading coefficient is negative, the derivative being zero corresponds to a maximum.

Maximum revenue: Substitute p = 123 back into the original equation: R(123) = -5(123)^2 + 1230(123) = $37,987.50.

Therefore, the company should set the camera price at $61.50 to achieve their maximum revenue of $37,987.50.

User Unwichtich
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