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 October 11th, 2015, 05:16 AM #1 Member   Joined: Aug 2015 From: Podgorica Posts: 61 Thanks: 0 Algebra A computer salesman finds that if he sells n computers, he can make a profit of $(200-4n) per computer. How many computers should he sell in order to maximize his profit? result: 25 Last edited by med1student; October 11th, 2015 at 05:47 AM.  October 11th, 2015, 05:29 AM #2 Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 If he sells n computers at \$(200 - 4n) profit per computer, he makes \$n(200 - 4n)= \$(200n - 4n^2). That is a quadratic function. Its graph is a parabola opening downward and the maximum value is at the vertex. Complete the square to find where the vertex of the graph is. Last edited by skipjack; October 11th, 2015 at 05:52 AM.
 October 11th, 2015, 05:58 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2215 The n-intercepts of the parabola occur for n = 0 and n = 50, so its vertex has n = 25.

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### a salesperson earns \$80 for every computer he sells. He earns an additional 250 for every 10 computers he sells that month. How many computers does he need to sell in order to earn 2340 in a month?

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