Optimization Problems

1. Optimization Problems • Assign a letter to each variable mentioned in the problem. Draw and label figure as needed. • Find an expression for the quantity to be optimized (Maximized or Minimized) • Use conditions to write expression as a function in one variable (note any domain restrictions). 4. Optimize the function using techniques for finding absolute maximum or absolute minimum.

2. Example An open box is formed by cutting identical squares from each corner of a 4 in. by 4 in. sheet of paper. Find the dimensions of the box that will yield the maximum volume. x 4 – 2x x x x 4 – 2x Why?

3. Critical points: The dimensions are 8/3 in. by 8/3 in. by 2/3 in. giving a maximum box volume of 4.74 in3.

4. There must be a local maximum here, since the endpoints are minimums. Why? A Classic Problem You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?

5. A Classic Problem You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?

6. 1 Write it in terms of one variable. • Find where the first derivative is equal to zero • or fails to exist. 3 Check the end points if necessary. To find the maximum (or minimum) value of a function:

7. Example: What dimensions for a one liter cylindrical can will use the least amount of material? Motor Oil We can minimize the material by minimizing the area. We need another equation that relates r and h: Too many variables! area of ends lateral area

8. Example : What dimensions for a one liter cylindrical can will use the least amount of material? area of ends lateral area

9. Notes: If the function that you want to optimize has more than one variable, use substitution to rewrite the function. If you are not sure that the extreme you’ve found is a maximum or a minimum, you have to check. Note, for the AP test, it is always best to justify! If the end points could be the maximum or minimum, you have to check.

10. EXAMPLE • The highway department is planning to build a picnic area for motorists along a major highway. It is to be rectangular with an area of 5000 square yards and is to be fenced off on the three sides not adjacent to the highway. What is the least amount of fencing that will be needed to complete the job?

11. SOLUTION… Let F denote the amount of fencing required. F = x + 2y The area is to be 5000, so xy = 5000, or y = 5000/x Now substitute y into F. The least amount of fencing needed is 200 yards.

12. The graph of F(x) = x + for x > 0.

13. EXAMPLE • A manufacturer can produce blank videotape cassettes at a cost of $2 apiece. The cassettes have been selling for $5 apiece, and at this price, consumers have been buying 4000 cassettes a month. The manufacturer is planning to raise the price of the cassettes and estimates that for each $1 increase in the price, 400 fewer cassettes will be sold each month. At what price should the manufacturer sell the cassettes to maximize profit?

14. SOLUTION… • Let x = the number of $1 increases in price. • Number of cassettes = 4000 – 400x • Profit per cassette= (5 – 2) + x = 3 + x • Total Profit = (# sold)(profit per cassette)= (4000 – 400x)(3 + x) for 0 < x < 10 • P'(x) = (4000 – 400x)(1) + (3 + x)(-400)] = 4000 – 400x – 1200 – 400x • = 2800 – 800x = 400(7 - 2x) • = 0 at x = 3.5

15. …SOLUTION • P'(x) = 0 at x = 3.5 • Since x = 3.5 is the only critical value on the interval 0 < x < 10 we can use the Second Derivative Test for Absolute Extrema. • P'(x) = 2800 – 800x • P"(x) = – 800 < 0 for all x in [0, 10] èAbsolute maximum occurs at x = 3.5 • The number of $1 increases in price is x = 3.5 • Maximum profit occurs when the cassettes are sold at $(5 + 3.5) = $8.50 each.

16. EXAMPLE • A cable is to be run from a power plant on one side of a river 900 meters wide to a factory on the other side, 3000 meters downstream. The cost of running the cable under the water is $5 per meter, while the cost over land is $4 per meter. What is the most economical route over which to run the cable?

17. Two reasonable choices for the variable x are shown below, with other unknowns labeled in terms of x. The hypotenuse of the triangle is found by using the Pythagorean Theorem. We will choose diagram (b) to make our calculations less complicated. SOLUTION… First, draw a diagram and label the given information and the unknowns. The cable must be run in a straight line from the power plant to some point P on the opposite bank.

18. …SOLUTION… Let C denote the cost of installing the cable. C = 5(# meters under water) + 4(# meters over land)

19. …SOLUTION The minimal installation cost is $14,700, which will occur if the cable reaches the opposite bank 1200 meters downstream.

20. EXAMPLE • A bus company will charter a bus that holds 50 people to groups of 35 or more. If a group contains exactly 35 people, each person pays $60. In large groups, everybody's fare is reduced by $1 for each person in excess of 35. Determine the size of the group for which the bus company's revenue will be greatest.

21. SOLUTION… • Let x = the number of people in excess of 35. • Number of people in the group = 35 + x • Fare per person = 60 – x • Revenue = (# people in group)(fare per person)R(x) =(35 + x)(60 – x) for 0 < x < 15 • R'(x) = (35 + x)(–1) + (60 – x)(1) = 25 – 2x • = 0 at x = 12.5

22. …SOLUTION… • R'(x) = 0 at x = 12.5 • Since x = 12.5 is the only critical value on the interval 0 < x < 15 we can use the second derivative test for absolute extrema. • R'(x) =25 – 2x • R"(x) = – 2 < 0 for all x in [0, 15] • Absolute maximum occurs at x = 12.5 • But x represents a number of people and must be a whole number, either 12 or 13. • R(12) = 2256 and R(13) = 2256 • Maximum revenue occurs when the group contains either 12 or 13 people in excess of 35, that is 47 or 48 people.

23. The revenue functionR(x) = (35 + x)(60 – x)

24. EXAMPLE • A bicycle manufacturer buys 6000 tires a year from a distributor. The ordering fee is $20 per shipment, the storage cost is 96 cents per tire per year, and each tire costs $5.75. Suppose that the tires are used at a constant rate throughout the year and that each shipment arrives just as the preceding shipment is being used up. How many tires should the manufacturer order each time to minimize cost?

25. SOLUTION… Let x denote the number of tires in each shipment and C(x) the corresponding total cost in dollars. Total cost = storage cost + ordering cost + purchase cost Since 6000 tires are ordered during the year and each shipment contains x tires, the number of shipments is 6000/x. Ordering cost = (cost per shipment)(# of shipments)= 20(6000/x) = 120,000/x Purchase cost = (total # of tires ordered)(cost per tire)= 6000(5.75) = 34,500 Storage cost = (average # of tires stored)(storage cost per tire)= (x/2)(0.96) = 0.48x

26. Inventory graphs: (a) actual inventory graph and (b) constant inventory of tires. …SOLUTION… The average number of tires in storage during the year is x/2, and the total yearly storage cost is the same as if x/2 tires were kept in storage for the entire year.

27. …SOLUTION… Total cost = storage cost + ordering cost + purchase cost Cost is minimized when 500 tires are ordered per shipment.

28. Total cost C(x) = 0.48x +