HW 12 Gagan Solutions

# HW 12 Gagan Solutions - Gauss Karl Friedric Homework 12 R...

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Gauss, Karl Friedric – Homework 12 – R Gompf – 57965 – Nov 03, 2004 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – ±nd all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points The marketing department of a company has determined that the cost (in dollars) of producing x units of a new product is given by C ( x ) = 400 + 240 x, while the demand equation for that product will be given by p = 1200 1 25 x. (i) What will be the marginal pro±t when x = 30, assuming all units produced can be sold? 1. P 0 (30) = \$962 . 60 2. P 0 (30) = \$957 . 60 correct 3. P 0 (30) = \$972 . 60 4. P 0 (30) = \$967 . 60 5. P 0 (30) = \$977 . 60 Explanation: The revenue R ( x ) from the sale of x units will be R ( x )= xp ( x ) = 1200 x 1 25 x 2 , so the pro±t on these units will be P ( x )= R ( x ) C ( x ) = 960 x 1 25 x 2 400 . The marginal pro±t will thus be given by P 0 ( x ) = 960 2 25 x. At x = 30, therefore, P 0 (30) = 957 . 60. 002 (part 2 of 2) 10 points (ii) At what production level will the marginal pro±t be zero? 1. x = 12250 units 2. x = 11750 units 3. x = 12750 units 4. x = 12000 units correct 5. x = 12500 units Explanation: The marginal pro±t will be zero when P 0 ( x ) = 960 2 25 x =0 , i.e. , when x = 12000 units. 003 (part 1 of 1) 10 points A company manufacturing cars at plant A in Detroit and plant B in St. Louis has found that its total cost function is C = x 2 +2 y 2 +2 xy + 190 x +30 y +50 when x cars are produced at plant A and y at plant B . What production level at plant A should the company set to minimize the costs of pro- duction when ±lling an order for 150 cars? 1. prod plant A =7 0 correct 2. prod plant A

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Gauss, Karl Friedric – Homework 12 – R Gompf – 57965 – Nov 03, 2004 2 3. prod plant A =6 0 4. prod plant A =7 5 5. prod plant A =6 5 Explanation: To determine the production level mimin- imzing costs we have to minimize C = x 2 +2 y 2 +2 xy + 190 x +30 y +50 , a function of two variables, subject to the restriction ( ) x + y = 150 . Using ( ) to eliminate y from C reduces the problem to optimization of a function C ( x )= x 2 + 2(150 x ) 2 +2 x (150 x ) + 190 x + 30(150 x )+50 = x 2 140 x + 49550 of one variable. After di±erentiation, C 0 ( x )=2 x 140 ,C 0 ( x )=2 . By the second derivative test for functions of one variable, C thus has a minimum value at x = 70, i.e. , to minimize production costs the company should set prod plant A =7 0 . 004 (part 1 of 1) 10 points A rancher wishes to build a fence to enclose a 3200 square yard rectangular ²eld. Along one side the fence is to be made of heavy duty material costing \$9 per yard, while the re- maining three sides are to be made of cheaper material costing \$3 per yard. Determine the least cost, C
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## This note was uploaded on 02/28/2010 for the course M 56200 taught by Professor Radin during the Fall '09 term at University of Texas.

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HW 12 Gagan Solutions - Gauss Karl Friedric Homework 12 R...

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