Optimization I.

1. Optimization I. Dr. Julia Arnold using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text

2. “Many real world applications call for finding the absolute maximum value or the absolute minimum value of a given function. For example, management is interested in finding what level of production will yield the maximum profit for a company; a farmer is interested in finding the right amount of fertilizer to maximize crop yield; a doctor is interested in finding the maximum concentration of a drug in a patient’s body and the time at which it occurs; and an engineer is interested in finding the dimension of a container with a specified shape and volume that can be constructed at a minimum cost.”1 In the next two sections we will be learning how to find the optimum shape and size of containers, or land, or whatever the problem bestows on us to find. 1 From Optimization I, section 4.4, page 345 Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text

3. First we need to talk about Absolute Extrema Definition: For a given function f(x), if for all x in the domain of f, then f(c ) is called the absolute maximum value of f. If for all x in the domain of f, then f(c ) is called the absolute minimum value of f. All f(x) values are above (0,0) in Figure 1. All f(x) values are below (0,2) in Figure 2. Figure 1 shows an absolute minimum for f(x) at the origin. Figure 2 shows an absolute maximum for f(x) at (0,2).

4. Theorem : If a function f is continuous on a closed interval [a,b], then f has both an absolute maximum value and an absolute minimum value on [a,b]. Continuous Function Absolute Maximum [ ] Closed Interval Absolute Minimum

5. Theorem : If a function f is continuous on a closed interval [a,b], then f has both an absolute maximum value and an absolute minimum value on [a,b]. This graph shows why the function has to be continuous. Goes to positive infinity, no absolute maximum. Function not continuous at x = 2 [ ] Closed Interval Goes to negative infinity, no absolute minimum.

6. Steps for finding the absolute extrema of a function f on a Closed Interval [a,b]. • Find the critical points of f that lie in (a,b) (open interval). • Compute the value of f at each critical point found in step 1 and also compute f(a) and f(b) • The absolute maximum value and absolute minimum value of f will correspond to the largest and smallest numbers, respectively, found in step 2.

7. Example 1: Find the absolute extrema of the function f(x) = x2 defined on the closed interval [-1,2]. Step 1: Find the critical points. f’(x) = 2x 0=2x X = 0 is a critical point Step 2: Calculate f(-1), f(0), and f(2) a=-1 and b=2, 0 is the critical point. f(-1)= 1 f(0) = 0 f(2) = 4 The point (0,0) is the lowest point on the interval, while (2,4) is the highest point on the interval. Thus (0,0) is the absolute minimum on [-1,2] and (2,4) is the absolute maximum on [-1,2]

8. Example 2: Find the absolute extrema for on the interval [0,3] Step 1: Step 2: Substitution into f(x) f(0),f(2),f(3) Finding Critical Points Why not –2/3? It’s not inside the closed interval. (0,4) is the absolute maximum (2,-4) is the absolute minimum

9. The Sonic Company’s total profit in dollars from manufacturing and selling x units of their loudspeaker system is given by How many units of the loudspeaker system must Sonic produce to maximize its profits. Solution: Find f(0), f(7500), f(20,000)

10. The largest profit occurs at the critical point x = 7500 and is $925,000.

11. A Biology Problem: When a person coughs, the trachea (windpipe) contracts, allowing air to be expelled at a maximum velocity. It can be shown that during a cough the velocity v of airflow is given by the function where r is the trachea’s radius (in centimeters), R is the trachea’s normal radius (in centimeters), and k is a positive constant that depends on the length of the trachea. Find the radius r for which the velocity of airflow is greatest. Since a trachea has a maximum size we can assume the closed interval [0,R] Solution:

12. Product Rule Remove grouping symbols Combine like terms and set equal to 0 Factor out r. Set each factor equal to 0. Solving for r. 2/3 R means that the radius is 2/3 the normal size of the trachea. Find v(0), v(2R/3), v(R )

13. minimum maximum minimum The airflow is greatest when the cough contracts the trachea to a radius of 2R/3 or when the contraction is approximately 1/3 of the Radius. Trachea at rest Trachea during a cough. 2/3R

14. Maximizing Profits The quantity demanded each month of the Walter Serkin recording of Beethoven’s Moonlight Sonata, manufactured by Phonola Record Industries, is related to the price/compact disc. The equation Where p denotes the unit price in dollars and x is the number of discs demanded, relates the demand to the price. The total monthly cost (in dollars) for pressing and packaging x copies of this classical recording is given by To maximize its profits, how many copies should Phonola produce each month?

15. Maximizing Profits Continued The equation p is the relationship between the price we are going to charge for the disc in terms of the quantity of discs demanded. So if 1000 discs are demanded, the price would be: Per disc To find the Revenue we multiply p by x. (price by number of items to be sold)

16. Now that we have both Revenue and Cost we can find the Profit function: Now that we have the profit function we can proceed to find its maximum value which will lead us to the number of discs we should make.

17. To find the maximum profit, you find the first derivative, set it equal to 0 and solve for x. X = 5000 discs made and sold will maximize the profits

18. If we look at the graph of the profit function we should see that x = 5000 is at the peak.

19. Self-Check See if you can do the 3 problems below. If you need help click on the appropriate link. 1. For the function a. Find the absolute extrema of f on [0,9] b. Find the absolute extrema of f(x). HintAnswerCompletely Worked 2. Find the absolute extrema of on [-2,1] HintAnswerCompletely Worked 3. The operating rate(expressed as a percentage) of factories, mines, and utilities in a certain region of the country on the tth day of the year 2000 is given by the function On which day of the first 250 days of 2000 was the manufacturing capacity operating rate highest? HintAnswerCompletely Worked End presentation

20. Hint: Step 1 Take the first derivative of the function, then set it equal to 0 and solve. Back to questions

21. Answers 1. a. (1,-1) absolute minimum and (9,3) absolute maximum b. (1,-1) is the absolute minimum, no absolute maximum Back to questions

22. Answers 2. (-1,0) absolute minimum and (-2,17) absolute maximum Back to questions

23. Answers 3. The 200 th day Back to questions

24. Complete Solution 1. a Step 1 Rewrite for the derivative Step 2 [0,9] and critical point x = 1 Use the power rule. Multiply by denominator Absolute minimum Absolute maximum Square both sides. 1. B The domain over the real numbers is (1,-1) is an absolute minimum for the entire function, but no max. Back to questions

25. Complete Solution 2. Step 1 Step 2 [-2,1] and critical point x = -1,0 Use the power rule. Factor GCF Set equal to 0 Absolute minimum Critical Values Absolute maximum Back to questions

26. Complete Solution 3. Step 1 Step 2 0,250] and critical point x = 200 Use the quotient rule. Only numerator can be 0 Factor and divide by 1200 on both sides Absolute maximum Solve for t2 On the 200 th day of 2000 the manufacturing capacity is highest. Back to questions

27. If you have any questions or comments about this presentation, please contact Dr. Julia Arnold at jarnold@tcc.edu