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October 11th, 2015, 05:16 AM   #1
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Angry 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.
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October 11th, 2015, 05:29 AM   #2
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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.
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October 11th, 2015, 05:58 AM   #3
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The n-intercepts of the parabola occur for n = 0 and n = 50, so its vertex has n = 25.
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