(a) Explanation Definition used: First derivative test: Suppose c to be a critical number for a function f . Assume that f is continuous on ( a , b ) and differentiable on ( a , b ) except at c , and c is the only critical number of f in ( a , b ) . f ( c ) is a relative maximum of f at c if f ′ ( x ) > 0 in the interval ( a , c ) and f ′ ( x ) < 0 in the interval ( c , b ) . f ( c ) is a relative minimum f at c if f ′ ( x ) < 0 in the interval ( a , c ) and f ′ ( x ) > 0 in the interval ( c , b ) . Relative maximum: A function is said to have a relative maximum f ( c ) at c if f ′ ( x ) > 0 in the interval ( a , c ) and f ′ ( x ) < 0 in the interval ( c , b ) . Relative minimum: A function is said to have a relative maximum f ( c ) at c if f ′ ( x ) < 0 in the interval ( a , c ) and f ′ ( x ) > 0 in the interval ( c , b ) . Calculation: The profit function is P ( x ) = ln ( − x 3 + 3 x 2 + 72 x + 1 ) . Obtain the derivative of the function as follows. P ′ ( x ) = ( ln ( − x 3 + 3 x 2 + 72 x + 1 ) ) ′ = 1 − x 3 + 3 x 2 + 72 x + 1 ⋅ ( − ( 3 x 2 ) + 3 ( 2 x ) + 72 ( 1 ) + 0 ) = − 3 x 2 + 6 x + 72 − x 3 + 3 x 2 + 72 x + 1 Set P ′ ( c ) = 0 and find the value of c . − 3 c 2 + 6 c + 72 − c 3 + 3 c 2 + 72 c + 1 = 0 − 3 c 2 + 6 c + 72 = 0 − 3 c 2 + 18 c − 12 c + 72 = 0 − 3 c ( c − 6 ) − 12 ( c − 6 ) = 0 That is, − 3 c ( c − 6 ) − 12 ( c − 6 ) = 0 ( − 3 c − 12 ) ( c − 6 ) = 0 − 3 c − 12 = 0 or c − 6 = 0 c = − 4 or c = 6 Therefore, the critical numbers are c = − 4 and c = 6 (b) To determine To find: The maximum profit for the given number of units.

Finite Mathematics and Calculus with Applications (10th Edition)

10th Edition

ISBN: 9780321979407

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 13.2, Problem 48E

(a)

To determine

To find: The number of units, q that gives maximum profit where the total profit is given by P(x)=ln(−x3+3x2+72x+1) .

(b)

To determine

To find: The maximum profit for the given number of units.

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Chapter 13.2, Problem 47E

Chapter 13.2, Problem 49E

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The number of units, q that gives maximum profit where the total profit is given by P ( x ) = ln ( − x 3 + 3 x 2 + 72 x + 1 ) .

Solution Summary: The author explains that the number of units sold for which the profit is the maximum is x=6.