Type your question here -0.4 points My Notes SEssCalcET2 2.1.0400, The numbers o

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-0.4 points My Notes SEssCalcET2 2.1.0400, The numbers of locations as of October 1 are given The number N of locations of a popular coffeehouse chain is given in the table Year 2004 2005 2006 2007 2008 8567 10,235 12,442 15,014 16,676 Find the average rate of growth between each pair of years. 2004 to 2006 locations/ year 2006 to 2007 locations/ year 2005 to 2006 locations/ year b) Estimate the instantaneous rate of growth in 2006 by taking the average of the last two rates of change in part (a locations/year Estimate the instantaneous rate of growth in 2006 by measuring the slope of the tangent line through (2005 10235) and (2007, 15014) (c) locations/year d) Estimate the instantaneous rate of growth in 2007 by measuring the slope of the tangent line through (2006 12442) and (2008, 16676) locations/year Compare the growth rates you obtained in part (c and (d). What can you conclude There is not enough information O The rate of growth is decreasing. The rate of growth is constant. O The rate of growth is increasing. Need Help? Read it Chat About It -0.4 points My Notes I SEssCalcET2 2.1.04 The cost (in dollars of producing x units of a certain commodity is C(x 4000 9x 0.05 when the production level is changed from x J 100 to the given value. Round your answers to the nearest a) Find the average rate of change of C with respect tox cent (i) x 104 per unit (ii) 101 per unit b) Find the instantaneous rate of change of C with respect to x when x 100. (This is called the marginal cost. per unit

Explanation / Answer

Task 6.

a)

2004-2006 Growth = (12442/8567)^(1/2) - 1 = 20.5%

2006-2007 Growth = (15014/12442) - 1 = 20.67%

2005-2006 Growth = (12442/10235) - 1 = 21.56%

b) Avereage in 2006 is = (20.67% +21.56%)/2 = 21.12%

c) N in 2006 that is passing those two points is = (15014+10235)/2 = 12625

Growth in 2006 is = N(2006)/ N(2005) = (12625/10235) - 1= 23.35%

d) N in 2008 that is passing those two points is = (16676+12442)/2 = 14559

Growth in 2007 is = N(2007)/ N(2006) = (14559/12442) - 1= 17.01%

The rate of growth is decresing (andwer 2)

Task 7

Please round the answer yourself because the description was cut off.

a) C(x) = 4000 + 9x + 0.05*x*x

C(100) = 4000 + 9* 100 + 0.05* 100* 100 = 4000 + 900 + 500 = 5400

C(104) = 4000 + 9*104 + 0.05*104*104 = 4000 + 936 + 540.8 = 5476.8

C(101) = 4000 + 9*101 + 0.05*101*101 = 4000 + 909 + 510.05 = 5419.05

i) average rate of change at x = 104 per unit is equal to (C(104)-C(100))/(104- 100) - 1 = (5476.8 - 5400) / (104 - 100) = 76.8 / 4 = 19.2

ii) average rate of change at x = 101 per unit is equal to (C(101)-C(100))/(101- 100) - 1 = (5419.05 - 5400) / (101 - 100) = 19.05/ 1 = 19.05

b) instantenious rate of change can be found by taking a a derivative

C'(x) = 9 + 0.1*x

C'(100) = 9 + 0.1*100 = 19

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