the average annual expenditures for a countrys consumer on dairy products can be approximated by the…

the average annual expenditures for a countrys consumer on dairy products can be approximated by the function g(x)=310.5 + 40.7 ln x, where x = 6 corresponds to 2006. (a) estimate the average expenditures for 2009 and 2011. (b) assuming that the model remains accurate, what is the first full - year in which expenditures on dairy products exceed $438? (a) to find the average expenditure in 2009, substitute for x in the function. (simplify your answer.)

the average annual expenditures for a countrys consumer on dairy products can be approximated by the function g(x)=310.5 + 40.7 ln x, where x = 6 corresponds to 2006. (a) estimate the average expenditures for 2009 and 2011. (b) assuming that the model remains accurate, what is the first full - year in which expenditures on dairy products exceed $438? (a) to find the average expenditure in 2009, substitute for x in the function. (simplify your answer.)

Answer

Explanation:

Step1: Determine the value of x for 2009

Since x = 6 corresponds to 2006, for 2009, the number of years from 2006 is 2009 - 2006=3. So x = 6 + 3=9.

Step2: Calculate the average expenditure for 2009

Substitute x = 9 into the function g(x)=310.5 + 40.7lnx. g(9)=310.5+40.7ln(9) First, find ln(9)≈2.1972. Then 40.7×2.1972 = 40.7×(2 + 0.1972)=40.7×2+40.7×0.1972 = 81.4+8.02604 = 89.42604. g(9)=310.5 + 89.42604=399.92604≈399.93.

Step3: Determine the value of x for 2011

The number of years from 2006 to 2011 is 2011 - 2006 = 5. So x=6 + 5 = 11.

Step4: Calculate the average expenditure for 2011

Substitute x = 11 into the function g(x)=310.5 + 40.7lnx. Find ln(11)≈2.3979. Then 40.7×2.3979=40.7×(2 + 0.3979)=40.7×2+40.7×0.3979 = 81.4+16.19453 = 97.59453. g(11)=310.5+97.59453 = 408.09453≈408.09.

Step5: Solve for x when g(x)>438

Set up the inequality 310.5 + 40.7lnx>438. Subtract 310.5 from both sides: 40.7lnx>438 - 310.5=127.5. Then lnx > \frac{127.5}{40.7}\approx3.1327. Exponentiate both sides using the base - e: x>e^{3.1327}. Since e^{3.1327}\approx22.97. Since x represents a year - related value and it must be an integer, x = 23. The year corresponding to x = 23 is 2006+(23 - 6)=2023.

Answer:

(a) For 2009, substitute 9 for x in the function, and the average expenditure is approximately $399.93. For 2011, substitute 11 for x in the function, and the average expenditure is approximately $408.09. (b) The first full year in which expenditures on dairy products exceed $438 is 2023.