SOLUTION: Maximum profit. A chain store manager has been told by the main office that daily profit, P, is related to the number of clerks working that day, x, according to the function P =

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Question 864999: Maximum profit. A chain store manager has been told by the main office that daily profit, P, is related to the number of clerks working that day, x, according to the function P = −25x2 + 300x. What number of clerks will maximize the profit, and what is the maximum possible profit?
Please demonstrate your solution for both the number of clerks that will maximize the profit as well as the total maximum profit possible, making sure to include all mathematical work and an explanation for each step.
Thank You
Matt
Found 2 solutions by josgarithmetic, stanbon:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
P=-25x%5E2%2B300x, daily profit for x clerks.

Finding the maximum P.
Factor the definition.
P=-25%28x%5E2-12x%29
P=-25x%28x-12%29
The maximum will happen in the middle of the x intercept values, so solve for these using -25x%28x-12%29=0.
x-intercepts are 0 and 12.

The midpoint of 0 and 12 is highlight%28x=6%29, for the maximum P.
Evaluate P for x=6.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Maximum profit. A chain store manager has been told by the main office that daily profit, P, is related to the number of clerks working that day, x, according to the function P = −25x^2 + 300x. What number of clerks will maximize the profit, and what is the maximum possible profit?
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Since P is a parabola opening downward from (-b/(2a),f(-b/(2a)))
# of clerks at the max is -b/(2a) = -300/(2(-25)) = 300/50 = 6
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Max Profit = P(6) = -25(36)+300(6) = -900+1800 = $900
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Cheers,
Stan H.
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