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4.7 Solving Max-Min Problems 1. Read3. Identify the known quantities and the unknowns. Use a variable. 2. Identify the quantity to be optimized. Write a model for this quantity. Use appropriate formulas. This is the primary function. 3. If too many variables are in the primary function write a secondary function and use it to eliminate extra variables. 4. Find the derivative of the primary function. 5. Set it equal to zero and solve. 6. Reread the problem and make sure you have answered the question. Figure 3.43: An open box made by cutting the corners An open box is to be made by cutting from a square sheet of tin. (Example 1) squares from the corner of a 12 by 12 inch sheet and bending up the sides. How large should the squares be cut to make the box hold as much as possible? Figure open box cuttingfrom the the corners An open3.43: box isAn to be made bymade cuttingbysquares corner of a from sheet ofbending tin. (Example 1) How large should the 12 by a 12square inch sheet and up the sides. squares be cut to make the box hold as much as possible? Maximize the volume V=lwh V = (12 – 2x) (12 – 2x) x =144x – 48x2 + 4x3 V = 144 – 96x + 12x2 = 12(12 –8x + x2) 12(12 –8x + x2) = 0 (6-x)(2-x) = 0 x = 6 or x = 2 V = -92 + 24x is negative at x = 2. There is a relative max. Box is 8 by 8 by 2 =128 in3. Figure 3.46: The graph of A = 2 r + 2000/r is concave up. 2 Minimizing surface area You have been asked to design a 1 liter oil can in the shape of a right cylinder. What dimensions will use the least material? Figure The tograph = 2oil r(1 liter + 2000/r is concave up. 3) can in the You have3.46: been asked designofa 1Aliter = 1000cm shape of a right cylinder. What dimensions will use the least material? 2 Minimize surface area 1000 S 2 r 2 rh where 1000 r h h 2 r 1000 2 2 1 S 2 r 2 r 2 2 r 2000r r 2 S 4 r 2000r 2000 4 r 2 0 r 3 4 r 2000 Use the 2nd derivative test 500 r3 5.42 to show values give local minimums. h 10.84 2 2 4.8 Business Terms C x = number of items p = unit price C = Total cost for x items R = xp = revenue for x items = average cost for x units P = R – C or xp - C The daily cost to manufacture x items is C = 5000 + 25x 2. How many items should manufactured to minimize the average daily cost. 5000 C 25x x C 5000 5000 x 2 x 2 25 25 0 5000 25x 0 2 x 200 14.14 14 items will minimize the daily average cost. 4.10 Old problem Given a function, find its derivative function derivative Inverse problem Given the derivative, find the function. . Find a function that has a derivative y = 3x2 The answer is called the antiderivative You can check your answer by differentiation Curves with a derivative of 3x2 Each of these curves is an antiderivative of y = 3x2 Antiderivatives Derivative y x n y kf ( x) y f ( x) g( x) Antiderivative x n 1 y C , n 1 n 1 y kf ( x) C y f ( x) g ( x) Find an antiderivative 3x4 5x2 x 2 x 1 x 3x 2 2 2 Find antiderivatives y 3 x 4 5 x 2 x 2 x5 x3 x 2 y 3 5 2x C 5 3 2 3 5 5 3 1 2 y x x x 2x C 5 3 2 Check by differentiating Find an antiderivative 1 x2 x 1 x 1 y 1 x x 2 x 3 1 2 2 y x 2x 2 C 3 y 3 x 2 2 2 1 2 9 x 12 x 4 4 x5 x3 y 9 12 4 x C 5 3 2 9 5 y x 4 x3 4 x C 5 Trigonometric derivatives d sin x cos x dx d cos x sin x dx d tan x sec2 x dx d cot x csc2 x dx d sec x sec x tan x dx d csc x csc x cot x dx Derivative Antiderivative y cos x y sin x C y sin x y cos x C y sec 2 x y tan x C y csc 2 x y cot x C y sec x tan x y sec x C y csc x cot x y csc x C