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5.3 Optimization of Exponential Functions

5.3 Optimization of Exponential Functions. Alex Hudecki , Margaret McConkey , Thye Phan. Recall: Derivative of y= e x. e is constant and its derivative is simply F’(x)= e x. For Example: a) f(x)= e x F’(x)= e x.

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5.3 Optimization of Exponential Functions

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  1. 5.3 Optimization of Exponential Functions Alex Hudecki, Margaret McConkey, ThyePhan

  2. Recall: Derivative of y= ex • e is constant and its derivative is simply F’(x)=ex • For Example: • a) f(x)=ex • F’(x)=ex • When e is raised to the power of f(x), its derivative is f’(X)=ex [g’(x)] • For Example: • b) f(x)= e2x-4 • f’(x)=(2)e2x-4

  3. Recall: Derivative of y=bx • The derivative of y=bxis simplybx times the natural log of b • For Example: • f(x)=3x • F’(x)=3x (ln3) • The derivative of y=bf(x) is bf(x) times the natural log of b times the derivative of f(x) • For Example: • b) f(x)=37x+3 • F’(x)=37x+3(ln3)(7)

  4. Algorithm For Finding Max and Min • Find domain • Take the derivative of the function • Set the derivative equal to 0 and solve • Sub x values and restrictions into F(x) to find max/min

  5. Example #1 A plant grows exponentially a certain number of centimeters in t days. It is not possible for this plant to grow longer than 800 days. Its growth is represented by the functionH(t)=40x2e-0.4t +30. Determine after how many days this plant reaches its maximum height. x ≤ 800 1) Find the derivative of H(t) H’(t)= 2(40)xe-0.4t+40x²(-0.4) e-0.4t H’(t)= 80xe-0.4t - 16x²e-0.4t Factor: H’(t)=e-0.4t(80x-16x²)

  6. 2) Set H’(t)=0 and solve for t • 0= e-0.4t(80x-16x²) e-0.4t will not equal 0, therefore 80x-16x² must equal 0 for this equation to be true 80x-16x²=0 16x (5-x) x=0 x=5 If x=0, that implies no growth, so x=5 Therefore, it will take 5 days to reach max height

  7. Example #2 The narwhal population was studied by a biologist in a specific geographical region. The biologist determined that the narwhal population followed this exponential function where the population P in hundreds is a function of time: P(t)=20/(1+3e-0.02t) The largest population the area can sustain is represented by the limit as t positive infinity. Determine the limit. Now find the maximum population using the algorithm for finding max and min values.

  8. a) As t approaches positive infinity, e-0.02t approaches 0. This is because e-0.02t = 1/(e0.02t) and as t gets larger the denominator gets larger. As the denominator of a fraction gets larger, what happens? Therefore, the maximum population is 20

  9. b) P(t)=20/(1+3e-0.02t) P’(t)={n’(x) (d(x)) + (d’(x) ( n(x))} P’(t)=(0)( (1+3e-0.02t)-(-0.06e-0.02t)(20) (1+3e-0.02t)2

  10. Example#3 Students are having a bake sale and determine that the number of people who are interested in the product after t days of marketing is represented by p(t)=0.7{1-e^(-0.2t)}. The neighbourhood where the students are having the bake sale has 10 million possible customers, and each customer who responds contributes $0.70 of revenue. However this does not take into consideration the cost of advertising which is $30 000 plus $5000 per day. Determine a function for profit with respect to time. How many days must the sale be advertised for max sales? Note: the budget is $200 000.

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