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Section 3.3 Optimization

Section 3.3 Optimization. Find the absolute extreme values of each function on the given interval. 1. F(x) = x 3 – 6x 2 + 9x +8 on [-1,2]. f is continuous and [-1,2] is a closed interval. Find the critical values f’ (x) = 3x 2 – 12x + 9 = 3(x 2 – 4x + 3)

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Section 3.3 Optimization

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  1. Section 3.3 Optimization Find the absolute extreme values of each function on the given interval. 1. F(x) = x 3 – 6x 2 + 9x +8 on [-1,2] f is continuous and [-1,2] is a closed interval. Find the critical values f’ (x) = 3x 2 – 12x + 9 = 3(x 2 – 4x + 3) The critical values are x =1 and 3. Eliminate 3, which is not in the interval CV: x = 1 f(1) = (1) 3 – 6(1) 2 + 9(1) +8 = 1 – 6 + 9 +8 = 12 maximum EP: x =21 f(2) = (1) 3 – 6(1) 2 + 9(1) + 8 = 8-24 + 18 + 8 = 10 x = -1 f(-1) = (-1) 3 – 6(-1) 2 + 9(-1) + 8 = -8 minimum

  2. Find the absolute extreme values of the following function on the given interval. 2. f(x) = x 4 + 4x 3 + 4x 2 on [-2,1] f is continuous and [-2,1] is a closed interval. Find the critical values. f’ (x) = 4x 3 + 12x 2 + 8x = 4x ( x 2 + 3x +2) = 4x ( x+2) (x+1) The critical values are x = 0, - 2 and -1. CV: x=0 f(0) = (0) 4 + 4(0) 3 + 4(0) 2= 0 minimum x = -2 f(-2) = (-2) 4 + 4(-2) 3 + 4(-2) 2 = 16 – 32 + 16 = 0 minimum x = -1 f(-1) = (-1) 4 + 4(-1) 3 + 4(-1) 2 = 1 – 4 + 4 = 1 EP: x = 1 f(1) = (1) 4 + 4(1) 3 + 4(1) 2 = 1 + 4 +4 = 9 maximum

  3. Find the absolute extreme values of each function on the given interval. 3. f(x) = 2x 5 – 5x 4 on [-1,3] f is continuous and [-1,3] is a closed interval. Find the critical values. f’(x) = 10x 4 – 20x 3 = 10x 3 (x - 2) the critical values are x = 0 and x = 2 CV: x = 0 f(0) = 2(0) 5 -5(0) 4 = 0 x = 2 f(2) = 2(32) -5(16) = - 16 minimum EP: x = -1 f(-1) = 2(-1) – 5(1) = -7 x = 3 f(3) = 2(243 – 5(81) = 81 maximum

  4. Find the absolute extreme values of each function on the given interval. 4. F(x) = 5 – x on [0,5] f is continuous and [0.5] is a closed interval. Find the critical values. f’ x = -1 Since f’ (x) = -1, there is no critical values. EP: x = 0 f(0) = 5-0 = 5 maximum x = 5 f(5) = 5 – 5 = 0 minimum

  5. Find the absolute extreme values of each function on the given interval. 5. f(x) = (x 2 -1) 2 on [-1, 1] f is continuous and [1,-1] is a closed onterval. Find the critical values. f’ (x) = 2(x 2 -1)(2x) = 4x(x 2 – 1) The critical values are x = -1, 0, and 1. CV: x = -1 f(-1) =[(-1) 2 - 1] 2 =0 minimum x = 1 f(1) = [(1) 2 – 1] 2 = 0 ,minimum x = 0 f(0) = [(0) 2 – 1] 2 = 1 maximum

  6. 6. Find the number of the interval [0,3] such that the number minus its square is : • as large as possible. • b. as small as possible • Let x = the number. We wish to consider f’(x) = x – x 2, which is continuous and [0,3] is closed. • Find the critical values. • f’(x) =1 – 2x • The critical value is x = ½ • CV: x = ½ f(½) = ½ -¼ = ¼ maximum • EP: x = 0 F(0) = 0 – 0 = 0 • x = 3 f(3) = 3 – 9 = -6 minimum • ½ • 3

  7. 7. Biomedical: Pollen Count The average pollen count in NYC on days x of the pollen season is P(x) = 8x – 0.2x 2 (for 0 < x < 40) On which days is the pollen count highest? P is the continuous and (0,40) is a open interval. Find the critical values. P (x) = 8 -0.4x = 04(20 – x) The critical value is x = 20. CV: x = 20 P(20) = 8(20) – 0.2(20) 2 = 80 The pollen count is highest on the twentieth day. OR use your calculator. and maximum.

  8. 8. GENERAL: Fuel Economy The fuel economy (in miles per gallon) of an average American midsized car is E(x) = -0.01x 2 + 0.62x + 10.4, where x is the driving speed (in miles per hour, 20 ≤ x ≤ 60) At what speed is the fuel economy greatest? E is continuous and [20,60] is a closed interval. Find the critical values. E’ (x) = -0.02x + 0.62 The critical value is x = 31 CV: x=31 E(31) = -0.01(31) 2 + 0.62(31) + 10.4 = 20.01 EP: x = 20 E(20) = -0.01(20) 2 + 0.62(20) + 10.4 = 18.8 x = 60 E(60) = -0.01(60) 2 + 0.62(60) + 10.4 = 11.6 Fuel economy is greatest at 31 mph. OR use your calculator. and maximum.

  9. 9. BUSINESS: Copier Repair A copier company finds that copies are x years old require on average, f)x) = 1.2x 2 - 4.7x + 10.8 repairs annually for 0 ≤ x ≤ 5. Find the year that requires the least repairs, rounding your answer to the nearest year. f is continuous and [0,5] is a closed interval. Find the critical values. f’(x) = 2.4x – 4.7 The critical value is 4.7/2.4 = 1.96 CV: f (1.96) = 6.20 EP: x = 0: F(0) = 1.2(0) 2 – 4.7(0) + 10.8 = 10.8 x = 5 f(5) = 1.2(5) 2 – 4.7(5) + 10.8 + 17.3 The second year requires the least repairs. OR use your calculator. and minimum.

  10. 10. GENERAL: Driving and AGE Studies have shown that the number of accidents a driver has, varies with the age of the drive, and is highest for very young and very old drivers. The number of serious accidents for drivers of age x during 2003 was approx. f(x) = 0.013x 2 – 1.25x + 48 for 16 ≤ x ≤ 85. Find the age that has the least accidents, rounding to the nearest year. OR use your calculator. and minimum.

  11. 11. ENVIRONMENTAL SCIENCE: Pollution Two chemical factories are discharging toxic waste into a large lake, and the pollution level at a point x, miles from factory A towards factory B is P(x) = 3x2 -72x + 576 parts per million (for 0 ≤ x ≤ 50.) Find where the pollution is the least. P is continuous and [0,50] is a closed interval. Find the critical values. P(x) = 6x – 72 The critical value is x = 12 CV: x = 12 P(12) = 3(12) 2 – 72(12) + 576 = 144 EP: x = 0 P(0) = 3(0) 2 – 72(0) + 576 = 576 x = 50 P(50) = 3(50) 2 – 72(50) + 576 = 4476 Pollution is the least 12 miles away from factory A towards factory B. OR use your calculator. and minimum.

  12. 12. BUSINESS: Maximum Profit Country Motorbikes Incorporation finds that is costs $200 to produce each motorbike, and that fixed costs are $1500 per day. The price function is P(x) = 600 – 5x, where p is the price (in dollars) at which exactly x motorbikes will be sold. Find the quantity Country Motorbikes should produce and the price it should charge to maximize profit. Also find maximum profit. C(x) = 200x +1500 p(x) = 600-5x R(x) = (600 – 5x)x = 600x – 5x 2 P(x) = R(x) – C(x) = 600x – 5x 2 – (200x + 1500) = - 5x 2 + 400x – 1500 To maximize profit, we consider P(x) = 0 P(x) = -10x + 400 = 0 x = 40 To maximize profit, 40 motorbikes should be sold. The price for 40 bikes is p(40) = 600 – 5(40) = $400 The maximum profit is P(40) = R(40) – C(40) = 400(40 – [200(40) + 1500] = $6500

  13. 12. BUSINESS: Maximum Profit Country Motorbikes Incorporation finds that is costs $200 to produce each motorbike, and that fixed costs are $1500 per day. The price function is P(x) = 600 – 5x, where p is the price (in dollars) at which exactly x motorbikes will be sold. Find the quantity Country Motorbikes should produce and the price it should charge to maximize profit. Also find maximum profit. P(x) = - 5x 2 + 400x – 1500 OR use your calculator to find maximum profit.

  14. 13. GENERAL: Parking Lot Design A company wants to build a parking lot along the side of one of its building using 800 feet or fence. If the side along the building needs no fence, what are the dimensions of the largest possible parking lot? Let x = the side perpendicular to the building and let y = the side parallel to the building. Since only three sides will be fences, 2x + y =800 y = 800 – 2x We with to maximize A = xy = x(800 -2x), so we take the derivative. A = x(800 -2x) = 800x = 2x 2 A’ = 800 -4x = 0 x = 200 y = 800 – 2x = 800 -400 = 400 The side parallel to the building is 400 feet and the side perpendicular to the building is 200 feet. Building Parking lot

  15. 13. GENERAL: Parking Lot Design A company wants to build a parking lot along the side of one of its building using 800 feet or fence. If the side along the building needs no fence, what are the dimensions of the largest possible parking lot? maximize A = x(800 -2x) on your calculator. Building Parking lot

  16. River 14. A farmer wants to make three identical rectangular enclosures along a straight river, as in the diagram shown below. If he has 1200 yards of fence, and if the sides along the river need no fence what should be the dimensions of each enclosure if the total are is to be maximized? y + x = 1200 xy = A y = 1200 – 4x A = (1200 – 4x)x = 1200 – 4x 2 A’ = 1200 – 8X = 8(150 –x) = 0 x = 150 y = 1200 – 4(150) = 600 Since there are three identical enclosures the four sides perpendicular to the river are 150 yards long and the side parallel to the river is 600 yards long.

  17. River 14. A farmer wants to make three identical rectangular enclosures along a straight river, as in the diagram shown below. If he has 1200 yards of fence, and if the sides along the river need no fence what should be the dimensions of each enclosure if the total are is to be maximized? Or use your calculator to maximize area A = (1200 – 4x)x

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