Answer :
Part 1: The point of maxima is 21.212
Part 2: x = 21.212 is approximately equal to the year 2022
Part 3: The percentage at that time is 16.37%
Part 4: The rate of change of the percentage at that time is 0.294
Part 1:
p(x) = −0.00022x^3 + 0.014x^2 − 0.0033x + 12.236
where x is the number of years since 2000, data from 0<=x<=50
Differentiating the function with respect to x and equating it to 0
p'(x) = -0.00066x^2 + 0.028x - 0.0033
We will find the critical points of p'(x)
p''(x) = dp'(x)/dx = -0.00132x + 0.028
Critical points of p'(x) lies at x where p''(x) = 0
Thus, -0.00132x + 0.028 = 0
x = 0.028/0.00132 = 21.212
Now, p'''(x) = -0.00132<0
x = 21.212 is a point of maxima of p'(x)
Therefore, the maximum possible values of x could lie at x = 21.212 as well as the extreme points of the domain p'(x), i.e. at x = 0 and 50
p'(0) = -0.00066(0)^2 + 0.028(0) - 0.0033 = -0.0033
p'(21.212) = -0.00066(21.212)^2 + 0.028(21.212) - 0.0033 = 0.2936
p'(50) = -0.00066(50)^2 + 0.028(50) - 0.0033 = -0.2533
Hence p'(x) attains maxima at x = 21.212
Therefore, p(x) increases most rapidly at x = 21.212
Part 2:
From the given condition in the question, x = 0 corresponds to the year 2000
Thus x = 21.212 is approximately equal to year 2022
Part 3:
The percentage at that time:
p(x) = −0.00022x^3 + 0.014x^2 − 0.0033x + 12.236
p(21.212) = -0.00022(21.212)^3 + 0.014(21.212)^2 - 0.0033(21.212) + 12.236
= 16.365
= 16.37%
Part 4:
Rate of change of the percentage at that time
p(x) = −0.00022x^3 + 0.014x^2 − 0.0033x + 12.236
p'(21.212) = -0.00066(21.212)^3 + 0.028(21.212) - 0.0033
= 0.2936
=0.294
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