   Chapter 10, Problem 52RE ### 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

# Per capita health care costs For the years from 2000 and projected to 2018, the U.S. per capita out-of-pocket cost for health care C (in dollars) can be modeled by the function C ( t ) =   0.118 t 3 −   2.51 t 2 +   40.2 t   +   677 where t is the number of years past 2000 (Source: U.S. Centers for Medicare and Medicaid Services).(a) When does the rate of change of health care costs per capita reach its minimum?(b) On a graph of C(t), what feature occurs at the t-value found in part (a)?

(a)

To determine

To calculate: The time at which the rate of change of health care costs per capita reach its minimum if the U.S. per capita cost for health care C can be modeled by C(t)=0.118t32.51t2+40.2t+677 where t is the number of years past 2000.

Explanation

Given Information:

The provided equation is,

C(t)=0.118t32.51t2+40.2t+677,

Where t is the number of years past 2000.

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.

Calculation:

Consider the provided equation is C(t)=0.118t32.51t2+40.2t+677,

In order to minimize the sensitivity, the following steps have to followed,

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

C=ddt(0.118t32.51t2+40.2t+677)=0.118ddt(t3)2.51ddt(t2)+40.2ddt(t)+ddt(677)=0.354t25.02t+40.2

The second derivative is as follows:

C=ddt(0.354t25

(b)

To determine

The feature that occurs in the t value obtained in part (a) if C(t)=0.118t32.51t2+40.2t+677.

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