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Absolute Extrema - Business Calculus

  1. A company's profit from a speculative investment over a 10-year period is modeled by P(t)=(t−2)2/3+10P(t) = (t-2)^{2/3} + 10 for t∈[0,10]t \in [0, 10]. What is the difference between the absolute maximum and absolute minimum profit over this 10-year period?
    Answer: $4.00 The derivative is P′(t)=23(t−2)−1/3=23t−23P'(t) = \frac{2}{3}(t-2)^{-1/3} = \frac{2}{3\sqrt[3]{t-2}}. The derivative is never zero. However, it is undefined at t=2t=2, making t=2t=2 a critical number. The interval is [0,10][0, 10]. We test the critical number and the endpoints: P(0)=(−2)2/3+10=43+10≈11.59P(0) = (-2)^{2/3} + 10 = \sqrt[3]{4} + 10 \approx 11.59. P(2)=(0)2/3+10=10P(2) = (0)^{2/3} + 10 = 10. P(10)=(8)2/3+10=(83)2+10=22+10=14P(10) = (8)^{2/3} + 10 = (\sqrt[3]{8})^2 + 10 = 2^2 + 10 = 14. The absolute maximum is 14 and the absolute minimum is 10. The difference is 14−10=414 - 10 = 4.
  2. A company's marginal revenue is given by R′(x)=140−2xR'(x) = 140 - 2x and its marginal cost by C′(x)=20+xC'(x) = 20 + x, where xx is the number of units produced. The company's fixed costs are $500, and its production is constrained to the interval x∈[0,50]x \in [0, 50]. What is the absolute maximum profit the company can achieve?
    Answer: $1900 First, find the profit function P(x)P(x). The marginal profit is P′(x)=R′(x)−C′(x)=(140−2x)−(20+x)=120−3xP'(x) = R'(x) - C'(x) = (140 - 2x) - (20 + x) = 120 - 3x. The profit function is the integral of marginal profit, including fixed costs: P(x)=∫(120−3x)dx=120x−32x2−500P(x) = \int (120 - 3x) dx = 120x - \frac{3}{2}x^2 - 500 (since fixed costs are subtracted from profit). To find the absolute maximum on [0,50][0, 50], find critical numbers by setting P′(x)=0P'(x) = 0, which gives 120−3x=0120 - 3x = 0, so x=40x=40. Now evaluate P(x)P(x) at the critical number and endpoints: P(0)=−500P(0) = -500, P(40)=120(40)−32(40)2−500=4800−2400−500=1900P(40) = 120(40) - \frac{3}{2}(40)^2 - 500 = 4800 - 2400 - 500 = 1900, and P(50)=120(50)−32(50)2−500=6000−3750−500=1750P(50) = 120(50) - \frac{3}{2}(50)^2 - 500 = 6000 - 3750 - 500 = 1750. The absolute maximum profit is $1900.
  3. A company's profit P(x)P(x) from manufacturing xx thousand units of a product is given by P(x)=−x3+9x2+120x−400P(x) = -x^3 + 9x^2 + 120x - 400. The company's production capacity is limited to 20 thousand units. At what production level within its capacity, x∈[0,20]x \in [0, 20], does the company experience its absolute minimum profit?
    Answer: 20 thousand units To find the absolute minimum profit on the closed interval [0,20][0, 20], we apply the Closed Interval Method. First, find the derivative of the profit function: P′(x)=−3x2+18x+120P'(x) = -3x^2 + 18x + 120. Set the derivative to zero to find critical numbers: −3(x2−6x−40)=0-3(x^2 - 6x - 40) = 0, which factors to −3(x−10)(x+4)=0-3(x-10)(x+4) = 0. The critical numbers are x=10x=10 and x=−4x=-4. Only x=10x=10 is in the interval [0,20][0, 20]. Now, evaluate the profit function at the critical number and the endpoints: P(0)=−400P(0) = -400, P(10)=−1000+900+1200−400=700P(10) = -1000 + 900 + 1200 - 400 = 700, and P(20)=−8000+3600+2400−400=−2400P(20) = -8000 + 3600 + 2400 - 400 = -2400. Comparing these values, the absolute minimum profit is −2400-2400, which occurs at a production level of x=20x=20 thousand units.
  4. The value of a volatile stock over a 6-month period (where xx is the number of months) is modeled by V(x)=2x3−15x2+24x+50V(x) = 2x^3 - 15x^2 + 24x + 50 for x∈[0,6]x \in [0, 6]. What is the absolute lowest value the stock reaches during this period?
    Answer: $34 To find the absolute minimum, find the derivative: V′(x)=6x2−30x+24V'(x) = 6x^2 - 30x + 24. Set V′(x)=0V'(x) = 0 to find critical numbers: 6(x2−5x+4)=06(x^2 - 5x + 4) = 0, which factors to 6(x−1)(x−4)=06(x-1)(x-4) = 0. Critical numbers are x=1x=1 and x=4x=4, both of which are in the interval [0,6][0, 6]. Evaluate V(x)V(x) at the critical numbers and endpoints: V(0)=50V(0) = 50. V(1)=2−15+24+50=61V(1) = 2 - 15 + 24 + 50 = 61. V(4)=2(64)−15(16)+24(4)+50=128−240+96+50=34V(4) = 2(64) - 15(16) + 24(4) + 50 = 128 - 240 + 96 + 50 = 34. V(6)=2(216)−15(36)+24(6)+50=432−540+144+50=86V(6) = 2(216) - 15(36) + 24(6) + 50 = 432 - 540 + 144 + 50 = 86. Comparing the values {50, 61, 34, 86}, the absolute lowest value is 34.
  5. The concentration of a drug in the bloodstream tt hours after injection is modeled by C(t)=8tt2+4C(t) = \frac{8t}{t^2 + 4} mg/L. A medical study requires monitoring the drug concentration from 1 hour to 6 hours after injection. During this monitoring period, what is the maximum concentration observed?
    Answer: 2.0 mg/L at t = 2 hours Find critical points using the quotient rule: C'(t) = [8(t² + 4) - 8t(2t)]/(t² + 4)² = [8t² + 32 - 16t²]/(t² + 4)² = (32 - 8t²)/(t² + 4)² = 8(4 - t²)/(t² + 4)². Setting C'(t) = 0: 4 - t² = 0, so t = ±2. Since we're considering t ≥ 0, the critical point is t = 2. On the interval [1, 6], evaluate C(t) at t = 1, 2, and 6. C(1) = 8(1)/(1 + 4) = 8/5 = 1.6. C(2) = 8(2)/(4 + 4) = 16/8 = 2.0. C(6) = 8(6)/(36 + 4) = 48/40 = 1.2. The maximum concentration is 2.0 mg/L at t = 2 hours.
  6. Based on the profit functions provided, which statement accurately compares the absolute maximum profits for the two models on the interval x∈[0,8]x \in [0, 8]?
    Answer: Model B has a higher absolute maximum profit than Model A. For Model A, PA′(x)=−2x+10=0P_A'(x) = -2x + 10=0 implies a critical number at x=5x=5. Test endpoints and critical number: PA(0)=5P_A(0)=5, PA(8)=21P_A(8)=21, PA(5)=30P_A(5)=30. Absolute max for A is 30. For Model B, PB′(x)=−3x2+18x−15=−3(x−1)(x−5)=0P_B'(x) = -3x^2 + 18x - 15 = -3(x-1)(x-5)=0 implies critical numbers at x=1,x=5x=1, x=5. Test endpoints and critical numbers: PB(0)=10P_B(0)=10, PB(8)=−46P_B(8)=-46, PB(1)=3P_B(1)=3, PB(5)=35P_B(5)=35. Absolute max for B is 35. Since 35>3035 > 30, Model B has a higher absolute maximum profit.
  7. An open-top box is to be constructed from a square piece of cardboard, 18 inches on a side, by cutting out an identical square from each of the four corners and folding up the sides. What is the absolute largest volume, in cubic inches, that such a box can have?
    Answer: 432 cubic inches Let xx be the side length of the square cut from each corner. The height of the box will be xx, and the length and width will both be 18−2x18-2x. The volume is V(x)=x(18−2x)2=4x3−72x2+324xV(x) = x(18-2x)^2 = 4x^3 - 72x^2 + 324x. The physical constraint requires x>0x > 0 and 18−2x>018-2x > 0, so the interval for xx is (0,9)(0, 9). We can analyze on the closed interval [0,9][0, 9]. The derivative is V′(x)=12x2−144x+324V'(x) = 12x^2 - 144x + 324. Setting V′(x)=0V'(x)=0 gives 12(x2−12x+27)=012(x^2 - 12x + 27)=0, which is 12(x−3)(x−9)=012(x-3)(x-9)=0. Critical numbers are x=3x=3 and x=9x=9. We test the endpoints and the valid critical number: V(0)=0V(0) = 0, V(9)=0V(9) = 0, and V(3)=3(18−6)2=3(12)2=432V(3) = 3(18-6)^2 = 3(12)^2 = 432. The absolute maximum volume is 432 cubic inches.
  8. A restaurant's daily profit from selling xx gourmet burgers is P(x)=−x3+45x2−600x+1000P(x) = -x^3 + 45x^2 - 600x + 1000. The restaurant operates within the production range of x∈[5,40]x \in [5, 40] burgers per day. What is the absolute minimum profit the restaurant can make in a day?
    Answer: $-15000 First, find the derivative: P′(x)=−3x2+90x−600P'(x) = -3x^2 + 90x - 600. Set P′(x)=0P'(x) = 0: −3(x2−30x+200)=0-3(x^2 - 30x + 200) = 0, which is −3(x−10)(x−20)=0-3(x-10)(x-20) = 0. The critical numbers are x=10x=10 and x=20x=20, both in the interval [5,40][5, 40]. We evaluate the profit at the critical numbers and endpoints: P(5)=−125+45(25)−600(5)+1000=−1000P(5) = -125 + 45(25) - 600(5) + 1000 = -1000. P(10)=−1000+4500−6000+1000=−1500P(10) = -1000 + 4500 - 6000 + 1000 = -1500. P(20)=−8000+18000−12000+1000=−1000P(20) = -8000 + 18000 - 12000 + 1000 = -1000. P(40)=−64000+45(1600)−600(40)+1000=−64000+72000−24000+1000=−15000P(40) = -64000 + 45(1600) - 600(40) + 1000 = -64000 + 72000 - 24000 + 1000 = -15000. The absolute minimum profit is $-15000.
  9. The concentration of a drug in a patient's bloodstream tt hours after injection is modeled by the function C(t)=100+40t−24t5/3C(t) = 100 + 40t - 24t^{5/3} for the first 8 hours, i.e., on the interval t∈[0,8]t \in [0, 8]. What is the absolute maximum concentration of the drug during this period?
    Answer: 116 To find the absolute maximum, we find the derivative of C(t)C(t): C′(t)=40−24(53)t2/3=40−40t2/3C'(t) = 40 - 24(\frac{5}{3})t^{2/3} = 40 - 40t^{2/3}. Set C′(t)=0C'(t) = 0 to find critical numbers: 40(1−t2/3)=040(1 - t^{2/3}) = 0, which gives t2/3=1t^{2/3} = 1, so t=1t=1. Now we evaluate C(t)C(t) at the critical number t=1t=1 and the endpoints t=0t=0 and t=8t=8. C(0)=100C(0) = 100. C(1)=100+40(1)−24(1)5/3=116C(1) = 100 + 40(1) - 24(1)^{5/3} = 116. C(8)=100+40(8)−24(85/3)=100+320−24(32)=420−768=−348C(8) = 100 + 40(8) - 24(8^{5/3}) = 100 + 320 - 24(32) = 420 - 768 = -348. The values are 100, 116, and -348. The absolute maximum concentration is 116.
  10. The population of a managed fish colony after tt months is modeled by P(t)=100te−t/2P(t) = 100te^{-t/2}, with the model being valid for the first year, t∈[0,12]t \in [0, 12]. What is the absolute maximum population the colony reaches during the first year?
    Answer: 200e−1200e^{-1} Using the product rule, the derivative is P′(t)=(100)e−t/2+(100t)e−t/2(−12)=100e−t/2(1−t2)P'(t) = (100)e^{-t/2} + (100t)e^{-t/2}(-\frac{1}{2}) = 100e^{-t/2}(1 - \frac{t}{2}). Setting P′(t)=0P'(t) = 0 gives 1−t2=01 - \frac{t}{2} = 0, so t=2t=2 is the critical number. We test this and the endpoints of the interval [0,12][0, 12]: P(0)=100(0)e0=0P(0) = 100(0)e^0 = 0. P(2)=100(2)e−2/2=200e−1≈73.6P(2) = 100(2)e^{-2/2} = 200e^{-1} \approx 73.6. P(12)=100(12)e−12/2=1200e−6≈2.97P(12) = 100(12)e^{-12/2} = 1200e^{-6} \approx 2.97. The absolute maximum population is 200e−1200e^{-1}.

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