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CHAPTER 3 SECTION 3.7 OPTIMIZATION PROBLEMS Applying Our Concepts • We know about max and min … • Now how can we use those principles? Use the Strategy • What is the quantity to be optimized? – The volume • What are the measurements (in terms of x)? • What is the variable which will manipulated to determine the optimum volume? 60” • Now use calculus x principles 30” Guidelines for Solving Applied Minimum and Maximum Problems Optimization Optimization Maximizing or minimizing a quantity based on a given situation Requires two equations: Primary Equation what is being maximized or minimized Secondary Equation gives a relationship between variables To find the maximum (or minimum) value of a function: 1 Write it in terms of one variable. 2 Find the first derivative and set it equal to zero. 3 Check the end points if necessary. 1. An open box having a square base and a surface area of 108 square inches is to have a maximum volume. Find its dimensions. 1. An open box having a square base and a surface area of 108 square inches is to have a maximum volume. Find its dimensions. Primary Secondary V x2y SA x 2 4 xy 108 x 2 4 xy 108 x 2 V (x) x 4 x 3 108 x x 4 4 27x 41 x 3 2 108 x 2 y 4x 2 108 6 y 4 6 3y V '( x ) 27 34 x 2 27 34 x 2 0 x 27 x 6 2 4 3 y x x Domain of x will range from x being as small as possible to x as large as possible. Largest Smallest (y is near zero) 2 (x is near zero) x 108 x 0 6, Intervals: 0,6 Test values: 1 V ’(test pt) V(x) inc dec Dimensions: 6 in x 6 in x 3 in rel max x 6 108 10 2. Find the point on f x x 2 that is closest to (0,3). Find the point on f x x 2 that is closest to (0,3). 2. Minimize distance Secondary Primary d x 0 y 3 d x y 3 2 2 yx 2 2 ***The value of the root will be smallest when what is inside the root is smallest. d x y 3 2 2 Intervals: d x x x 3 d x x 2 x 4 6x 2 9 2 2 d x x 4 5x 2 9 d ' x 4x 3 10x 4 x 3 10 x 0 2x 2x 2 5 0 x 0 x 0 2x 2 5 0 x 52 2 2 , Test values: d ’(test pt) d(x) 5 2 3 5 2 ,0 1 dec 5 2 , inc dec inc rel max 5 2 5 5 2 2 , 5 2 3 rel min x 0, 1 rel min x 5 2 5 5 2 2 , 2. A rectangular page is to contain 24 square inches of print. The margins at the top and bottom are 1.5 inches. The margins on each side are 1 inch. What should the dimensions of the print be to use the least paper? 2. A rectangular page is to contain 24 square inches of print. The margins at the top and bottom are 1.5 inches. The margins on each side are 1 inch. What should the dimensions of the print be to use the least paper? Primary A x 2 y 3 A( x ) x 2 24x 3 24 3 x 48x 6 1 3 x 48 x 30 A '( x ) 3 48 x2 2 Secondary 1.5 xy 24 y 24x y 24 in 1 24 4 Print dimensions: 6 in x 4 in Page dimensions: 9 in x 6 in y 1 x 1.5 y 6 3 x 48 x2 crit #'s: x 0, 4 y 3 2 x2 Smallest Largest (x is near zero) (y is near zero) x 24 x 0 Intervals: 0,4 4,24 Test values: 1 10 A ’(test pt) A(x) dec inc rel min x4 1. Find two positive numbers whose sum is 36 and whose product is a maximum. 1. Find two positive numbers whose sum is 36 and whose product is a maximum. Primary P xy P x x 36 x P '( x ) 36 2 x Test values: 0,18 18,36 20 1 P ’(test pt) P(x) inc dec rel max x 18 x y 36 y 36 x y 36 18 18 18,18 36 2x 0 x 18 Intervals: Secondary 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? A x 40 2 x x x 40 2x w x l 40 2x w 10 ft l 20 ft A 40 x 2 x 2 A 40 4x 0 40 4x 4x 40 x 10 There must be a local maximum here, since the endpoints are minimums. 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? A x 40 2 x x x 40 2x w x l 40 2x w 10 ft l 20 ft A 40 x 2 x 2 A 40 4x 0 40 4x 4x 40 x 10 A 10 40 2 10 A 10 20 A 200 ft 2 Example 5: What dimensions for a one liter cylindrical can will use the least amount of material? We can minimize the material by minimizing the area. We need another equation that relates r and h: V r 2h 3 1 L 1000 cm 1000 r 2 h 1000 h 2 r A 2 r 2 2 rh area of ends lateral area 1000 A 2 r 2 r r2 2 2000 A 2 r r 2 2000 A 4 r 2 r Example 5: What dimensions for a one liter cylindrical can will use the least amount of material? A 2 r 2 2 rh V r 2h 3 1 L 1000 cm 1000 r 2 h 1000 h 2 r 1000 5.42 2 area of ends lateral area 1000 2 A 2 r 2 r r2 2000 4 r 2 r 2000 4 r 3 500 2000 A 2 r r 2 h h 10.83 cm r3 500 2000 A 4 r 2 r r 2000 0 4 r 2 r r 5.42 cm 3 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. If the end points could be the maximum or minimum, you have to check. Example #1 • A company needs to construct a cylindrical container that will hold 100cm3. The cost for the top and bottom of the can is 3 times the cost for the sides. What dimensions are necessary to minimize the cost. 2 r V r h h SA 2rh 2r 2 Minimizing Cost V r 2h 100 r 2 h SA 2rh 2r 2 100 SA 2r 100 r 2 2r 200 SA 2r 2 r Domain: r>0 r 2 2 h 200 C (r ) 6r 2 r C(r ) 200 r 2 12r Minimizing Cost C(r ) 200 12r r2 200 0 2 12r r 200 r2 12r r 3 50 3 100 3 12 r C (1.744) 0 Concave up – Relative min ------ +++++ 0 1.744 C' changes from neg. to pos. Rel. min 200 12r 3 200 12 C (r ) r3 1.744 100 h r h 10.464 The container will have a radius of 1.744 cm and a height of 10.464 cm 2