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11_Section1.2_new

11_Section1.2_new - ﬁg American University of Sharjah MTH...

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Unformatted text preview: ﬁg; American University of Sharjah MTH 101 — Mathematics For Business 1 Fall 2007 11th Edition solutions to the selected problems of section 12 Solutions by Omar Adnan Al Ismail Question 5 9: Fixed costs = \$124/day Variable costs = \$0.12/doughnut Number of doughnuts produced = x A) Find the daily cost function C(x). Total cost = Fixed costs + Variable costs = PC + (VC per doughnut. Number of doughnuts produced) = 124 + 0.12 x B) Find the number of doughnuts produced if the total cost was \$250 C(x) = 250 250 = 124 + 0.12 x 250 - 124 = 0.12 x X = 1050 doughnuts can be produced for the cost of \$250 Question 61 (Graded): Number of _olf clubs/ da _ \$7,647 \$9,147 A)Find the cost function C00 333’ C(x2)— C(xl) 9147—7547 1500 S Ope Ax xz—xl 100-80 20 5 C(x) = 75 x + b 7647 = 75 (80) + b B = 1647 C(x) = 75 x + 1647 B) Graph the function C(x) where 0 S x S 200 17500 15000 12500 10000 7500 5000 2500 IIJIIIIIII: 25 so 15 100 125 150 115 200 225 250 Question 63: Scum- Price R x \$112 \$175 \$238 A)Find the Selling Price Function R(x) _ _ Ay _ chz)— R(x1) _ 238—112 _ m 510136 Ax xz-xl 175—85 _ 1'4 R(x) = 1.4 x + b 112 = 1.4 (85) + b B=-7 C(x) = 1.4x — 7 B)Find the cost of a product with a selling price of \$185. C(x) = 185 185 = 1.4x - 7 185 + 7 = 1.4x x x \$137 Question 65 (Graded): Tractor’s Value V t _ \$157,000 \$82,000 A)Find the tractor’s value function in terms of t years. 111 ___ slope = n_y = V(t2)— V(t1) = 232,000—157,000 = _7500 Ax tz—tl 10—0 V(t) = —7500 t + b y - intercept = 157,000 V(t) = —7500 t + 157000 B) Find the tractor’s value after 6 year. V(6) = —7500 (6) + 157000 = \$112,000 C) After how many years will the tractor has a value less than \$7000? V(t) = 7000 7000 = —7500 t + 157000 t= 11.6 years D)Graph V(t) where 0 S t S 20 Y 350000 — — aauooo— — zsnnon— — zouooo— — 159mm ‘— — 108000 '- L 50000 —- — Question 75: 7.500 7.900 7.900 7,800 0 Find the price-supply equation m = slope = A—y = —P(x2)" P("1)= —2'37'”8 = 0.000225 Ax x2 -x1 7900—7500 P(x) = 0.000225 x + b 2.28 = 0.000225 (7500) + b B = 0.5925 P(x) = 0.000225 x + 0.5925 0 Find the price-demand equation Ay _ P(x2)— p(x1) _ 2.37—2.28 m = slope =————— = —0.0009 P(x) = —0.0009 x + b 2.28 = —0.0009 (7900) + b B = 9.39 P(x) = — 0.0009 x + 9.39 0 Find the equilibrium point. 0.000225 x + 0.5925 = —0.0009 x + 9.39 x = 7820, p(x) = 2.352 The point: (7820, 2.352) 0 Graph the supply and demand functions 5‘! ‘ '|"'l"l"'l"' 4.5—- X . I . . . l . . . I . . . i . . . I . . . . . . | . . '. 2000 4000 0000 3000 10000 12000 14000 ...
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