a. Explanation Given: The growth rate function is R ( M ) = − a M ln ( M K ) . Calculation: Consider the growth function R ( M ) = − a M ln ( M K ) . Substitute a = 1 and K = 4 in the function, R ( M ) = − a M ln ( M K ) = − ( 1 ) M ln ( M 4 ) = − M ln ( M 4 ) Use online graphing calculator to sketch the function R ( M ) = − M ln ( M 4 ) as shown below in Figure 1. From Figure 1, it is observed that the function R ( M ) = − M ln ( M 4 ) is positive value in the interval [ 0 , 4 ] . The function R ( M ) = − M ln ( M 4 ) is maximum for M = 1.471 . Given that the growth function is R ( M ) = − a M ln ( M K ) . Differentiate the function with respect to M . R ′ ( M ) = d d M ( − a M ln ( M K ) ) = − a ln ( M K ) ( 1 ) − a M k × 1 M K = − a ln ( M K ) − a = − a ( ln ( M K ) + 1 ) Set the above derivative to zero to obtain the critical point. − a ( ln ( M K ) + 1 ) = 0 ln ( M K ) + 1 = 0 ln ( M K ) = − 1 Take exponential on both the sides. e ln ( M K ) = e − 1 M K = 1 e M = K e Thus, the critical point is M = K e b. To determine To solve: The given initial value problem and graph the solution for a = 1 , K = 4 and M 0 = 1 . c. To determine To explain: The meaning of K .

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 7.9, Problem 66E

To determine

To sketch: The graph of the growth rate function R(M)=−aMln(MK) when a=1 and K=4 at which value of M the growth rate is positive and maximum.

To determine

To solve: The given initial value problem and graph the solution for a=1, K=4 and M0=1.

To determine

To explain: The meaning of K.

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Chapter 7.9, Problem 65E

Chapter 7.9, Problem 67E

Chapter 7 Solutions

Calculus: Early Transcendentals (2nd Edition)

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The graph of the growth rate function R ( M ) = − a M ln ( M K ) when a = 1 and K = 4 at which value of M the growth rate is positive and maximum.

Solution Summary: The author illustrates the graph of the growth rate function R(M)=-aM

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To sketch: The graph of the growth rate function R(M)=−aMln(MK) when a=1 and K=4 at which value of M the growth rate is positive and maximum.

To solve: The given initial value problem and graph the solution for a=1, K=4 and M0=1.

To explain: The meaning of K.

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Chapter 7 Solutions