   Chapter 10, Problem 38RE ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Productivity—diminishing returns Suppose the productivity P of an individual worker (in number of items produced per hour) is a function of the number of hours of training t according to P ( t ) = 5 + 95 t 2 t 2 + 2700 Find the number of hours of training at which the rate of change of productivity is maximized. (That is, find the point of diminishing returns.)

To determine

To calculate: The number of hours of training at which the rate of change of change of productivity is maximized, if suppose the productivity P of an individual worker(in terms of items produced per hour) is a function of the number of hours of training t according to P(t)=5+95t2t2+2700.

Explanation

Given Information:

The provided equation is P(t)=5+95t2t2+2700, where the number of hours of training t.

Formula Used:

The critical values are the only values at which the graph can have turning points, the derivative cannot change sign anywhere except at the critical value.

This, in an interval between two critical values, the sign of the derivative at any value in the interval will be the sign of the derivative at all values in the interval.

As per the First Derivative Test,

The first derivative of the function is evaluated. The first derivative is made equal to zero in order to get the critical points.

The values of the critical values are kept inside the original function which gives the critical points. The intervals of the values of x are then evaluated for the relative maximum and minimum.

The rate is the first derivative.

Calculation:

Consider the provided equation P(t)=5+95t2t2+2700,

The critical values are the only values at which the graph can have turning points, the derivative cannot change sign anywhere except at the critical value.

Hence, there will no change in the values of critical values as in the derivative graph.

Take out the first derivative of the equation by the power rule,

P(t)=(t2+2700).190t95t2.2t(t2+2700)2=513000t(t2+2700)2

The second derivative of the equation is as follows:

P(t)=513000t(t2+2700)2P(t)=(t2+2700)2.513000513000t.2(t2+2700).2t(t2+2700)4

Simplify the equation further:

P(t)=(t2+2700)2.513000513000t.2(t2+2700).2t(t2+2700)4=513000(t2+2700)[(t2+2700)4t2](t2+2700)4=513000(27003t2)(t2+2700)3

Put the value of second derivative as 0,

513000(27003t2)(t2+2700)3=0(27003t2)=0t=900t=±30

Test the derivative at the left and right sides of all the critical points

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