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Lesson 4-7

Lesson 4-7. Optimization Problems. Ice Breaker. Using a blank piece of paper: Find the extrema (maximum) of A(w) = 2400w – 2w². A’(w) = 2400 – 4w. A’(w) = 0 = 2400 – 4w 4w = 2400 w = 600. A’’(w) = – 4 < 0 for all w so absolute maximum of A = 720,000.

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Lesson 4-7

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  1. Lesson 4-7 Optimization Problems

  2. Ice Breaker Using a blank piece of paper: Find the extrema (maximum) of A(w) = 2400w – 2w² A’(w) = 2400 – 4w A’(w) = 0 = 2400 – 4w 4w = 2400 w = 600 A’’(w) = – 4 < 0 for all w so absolute maximum of A = 720,000

  3. Objectives • Use derivatives to solve optimization problems • Find minimums and maximums

  4. Optimization Example Govt Fencing Farmer’s Field A farmer whose property borders I-81 wants to maximize the size of the pasture for cattle he can create with his limited resources. He can only afford to buy materials for 2400 feet of fencing. How big can he make his pasture? Primary equation (min or max) : A = l• w Secondary equation: l + 2w = 2400 so l = 2400 – 2w A = (2400 – 2w) • w = 2400w – 2w² (0 < w < 1200) dA --- = 2400 – 4w = 0 when 4w = 2400 or w = 600 ft dw Checking endpoints make no sense here, no area without either w or l A = 600 • (2400 – 1200) = 720,000 sq ft

  5. Procedures for Solving Optimization Problems • Assign variables to all given quantities to be determined. When feasible, make a sketch. Label the picture with the quantities. • Find a primary expression for the quantity to be optimized. • Reduce this primary equation to one having a single independent variable – this may involve the use of “secondary” equations (restrictions) relating the independent variable of the original equation. • Determine the domain for the independent variable, usually an interval. • Find the critical points and end points. • Use the techniques of calculus to determine the desired maximum or minimum value. • Answer the question (including units).

  6. Now take out your 4.7 notes

  7. Equations in Optimization Problems • Primary Equation • Contains the variable (dependent) for which we are trying to find the min or the max of • Secondary Equation • Contains a relationship (equation) that we can use to eliminate an independent variable in the primary equation  reducing it to only one variable!

  8. Example 1 A cone with slant height of 6 inches is to be constructed. What is largest possible volume of such a cone? Primary Equation: V = ⅓πr²h Secondary Equation: l ² = r² + h² l – slant height h – height l h r V = ⅓πr²h V = ⅓π(36 – h²)h 36 = r² + h² 36 - h² = r² dV/dh = ⅓π(36 – 3h²) V’ = ⅓π(36 – 3h²) = 0 when 3h² = 36 h² = 12 h = ± 23 (the minus has no meaning) V’’ = ⅓π(0 – 6h) concave down for all h>0 so a maximum V = ⅓π(36 – h²)h = ⅓π(36 – 12)23 = 163π= 87.06 cu inches

  9. Example 2 Find the points on the graph of x² - 4y² = 4 that are closest to the point (5,0). Primary Equation: D² = (x – 5)² + (y – 0)² Secondary Equation: x² - 4y² = 4 D² = (x – 5)² + (y)² D² = (x – 5)² + ¼(x² - 4) D =  (x – 5)² + ¼(x² - 4) x² - 4y² = 4 ¼(x² - 4) = y² dD/dx = 5(x – 4) / 2 (5x² - 40x + 96)1/2(by your calculator) D’ = 0 when (x – 4) = 0 denominator can’t = 0 x = 4 y = ±3 (from x² - 4y² = 4) D’’ (via calculator) : 40 / (5x² - 40x + 96)3/2 > 0  concave up min D = 2 when x = 4 so (4,3) and (4,-3) are points closest

  10. Example 2 (simpler way) Find the points on the graph of x² - 4y² = 4 that are closest to the point (5,0). Primary Equation: D² = (x – 5)² + (y – 0)² Secondary Equation: x² - 4y² = 4 D² = (x – 5)² + (y)² D² = (x – 5)² + ¼(x² - 4) Let z = D² x² - 4y² = 4 ¼(x² - 4) = y² dz/dx = 2(x – 5)(1) + ¼ (2x – 0) = 2x – 10 + ½ x = 2.5x – 10 z’ = 0 when 2.5x – 10 = 0 2.5x = 10 x = 4 y = ±3 z’’ (easy to deal with) : 2.5 > 0  concave up always  min D = 2 when x = 4

  11. Example 3 A flyer is to contain 50 square inches of printed matter, with 4-inch margins at the top and bottom and 2-inch margins on each side. What dimensions for the flyer will use the least paper? Primary Equation: A = (h + 8)(w + 4) Secondary Equation: 50 = h w A = 50 A = (h + 8)(w + 4) A = (h + 8)(50/h + 4) A = 4h + 400/h + 82 50 = h w 50 / h = w dA/dh = 4 – 400 / h² A’ = 4 – 400 / h² = 0 when 400 / h² = 4 h² = 10 h = ± 10 (the minus has no meaning) A’’ = 800h-3 concave up for all h>0 so a minimum Dimensions: h + 8 = 11.1623 inches, w + 4 = 50/10 + 4 = 19.8114 inches

  12. Example 4 An open box having a square base is to be constructed from 108 square inches of material. What dimensions will produce a box with maximum volume? Primary Equation: V = h∙s² Secondary Equation: 108 = s² + 4sh h s V = h∙s² V = (108 – s²)∙s/4 V = 27s – ¼ s³) 108 = s² + 4sh h = (108 – s²) / (4s) s dV/ds = 27 – ¾s² V’ = 27 – ¾s² = 0 when 27 = ¾s² 36 = s² s = ± 6 (the minus has no meaning) V’’ = -3/2s concave down for all s>0 so a maximum Dimensions: s= 6 inches, h = (108 – s²) / (4s) = 3 inches

  13. Example 5 Find two positive numbers that minimize the sum of twice the first number plus the second number, if the product of the two numbers is 288. Primary Equation: S = 2x + y Secondary Equation: 288 = x ∙ y S = 2x + y S = 2x + 288 / x 288 = x ∙ y 288 / x = y dS/dx = 2 – 288 / x² S’ = 2 – 288 / x² = 0 when 2 = 288 / x² 288 = 2x² s = ± 12 (the minus has no meaning) S’’ = 576 / x³ concave up for all x > 0 so a minimum Numbers: x = 12, y = 288 / x = 24

  14. Example 6 Two posts, one 28 feet tall and the other 12 feet tall, stand 30 feet apart. They are to be held by wires, attached to a single stake, running from ground level to the tops of the poles. Where should the stake be placed to use the least wire? Primary Equation: W = [28² + (30-x)²] + [x² + 12²] 28 12 x 30-x Secondary Equation: None (if you used 2 variables 30 = x + y) 30 dW/dx = (x – 30) / (x²-60x+1684) + x / (x² + 144) W’ = 0 when x = 9 (using the solve function on the calculator!) W’’ (9) = 32/525 concave up for x = 9 so a minimum Location: 9 feet away from smaller post

  15. Example 7 Bill has 4 feet of wire and wants to form a circle and a square from the wire. How much of the wire should he use on the square and how much should be used on the circle to enclose the maximum total area? Primary Equation: A = πr² + s² Secondary Equation: 4 = 2πr + 4s A = πr² + s² A = πr² + [(4 – 2πr)/4]² A = (π²/4 + π)r² - πr +1 4 = 2πr + 4s s = (4 - 2πr) / 4 dA/dr = 2(π²/4 + π)r - π A’ = 2(π²/4 + π)r - π when 2(π²/4 + π)r = π r = π / [2(π²/4 + π)] r = 0.28 feet or 3.36 inches A’’ = 2(π²/4 + π) concave up for all r so a minimum! Check endpoints! Dimensions: s= 12 inches, r = 0

  16. Summary & Homework • Summary: • Optimization problems usually involve two equations: a primary that relates independent variables to the item we are trying to optimize and a secondary equation that relates the two independent variables • Use Calculus (closed interval method) to determine and evaluate the extrema • Homework: • pg 336-341: 3, 4, 7, 9, 15, 23, 30, 33

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