Explanation Given: The function is R ( x ) = 4 27 ( − x 3 + 66 x 2 + 1050 x − 400 ) , where R ( x ) is the revenue calculated in thousand dollars and x is the amount spent on advertising. The range of x is 0 ≤ x ≤ 25 . Calculation: Evaluate R ′ ( x ) as follows. R ′ ( x ) = d d x ( 4 27 ( − x 3 + 66 x 2 + 1050 x − 400 ) ) = 4 27 ( − 3 x 2 + 66 ( 2 x ) + 1050 ( 1 ) ) = 4 27 ( − 3 x 2 + 132 x + 1050 ) Thus, R ′ ( x ) = 4 27 ( − 3 x 2 + 132 x + 1050 ) . The second derivative R ″ ( x ) can be obtained from R ′ ( x ) = 4 27 ( − 3 x 2 + 132 x + 1050 ) as follows. R ′ ′ ( x ) = d d x ( 4 27 ( − 3 x 2 + 132 x + 1050 ) ) = 4 27 ( − 3 ( 2 x ) + 132 ( 1 ) ) = 4 27 ( − 6 x + 132 ) Thus, the value of R ″ ( x ) is 4 27 ( − 6 x + 132 ) . The point of diminishing returns occurs at the inflection point or R ″ ( x ) = 0 . So, equate R ″ ( x ) to 0. 4 27 ( − 6 x + 132 ) = 0 − 6 x + 132 = 0 6 x = 132 x = 132 6 x = 22 To test the concavity at x = 22 divide the domain into two intervals ( 0 , 22 ) and ( 22 , 25 ) . Take a test point in each of the intervals and find the value of R ″ ( x ) . Let x = 20 be a test point in the interval ( 0 , 22 )

Finite Mathematics and Calculus with Applications

1st Edition

ISBN: 9781323188361

Author: Margaret Lial

Publisher: Pearson Education

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Chapter 13.3, Problem 73E

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To find: The point of diminishing returns (x,y) for the given function.

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Chapter 13.3, Problem 72E

Chapter 13.3, Problem 74E

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Finite Mathematics and Calculus with Applications

Ch. 13.1 - Find where the function is increasing and...Ch. 13.1 - Prob. 2YT Ch. 13.1 - Prob. 3YT Ch. 13.1 - Prob. 4YT Ch. 13.1 - Prob. 1WE Ch. 13.1 - Prob. 2WE Ch. 13.1 - Prob. 3WE Ch. 13.1 - Prob. 4WE Ch. 13.1 - Prob. 5WE Ch. 13.1 - Prob. 6WE

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The point of diminishing returns ( x , y ) for the given function.

Solution Summary: The author explains that the point of diminishing returns is (x,y) for the given function.

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