The monthly profit for a small company that makes long-sleeve T-shirts depends on the price per shirt. If the price is too high, sales will drop. If the price is too low, the revenue brought in may not cover the cost to produce the shirts. After months of data collection, the sales team determines that the monthly profit is
approximated by f(p) = -50p2 + 2050p-20,700, where p is the price per shirt and./ (p) is the monthly profit based on that price.
(a) Find the price that generates the maximum profit.
(b) Find the maximum profit.
(c) Find the price(s) that would enable the company to break even. If there is more than one price, use the "and" button.
To find the price that generates the maximum profit, we need to find the critical points of the profit function. This can be done by finding the derivative of f(p) and setting it equal to 0.
f'(p) = -100p + 2050
Now, set f'(p) = 0 and solve for p:
-100p + 2050 = 0
-100p = -2050
p = 20.50
Therefore, the price that generates the maximum profit is $20.50 per shirt.
To find the maximum profit, plug the price back into the profit function:
f(20.50) = -50(20.50)^2 + 2050(20.50) - 20,700
f(20.50) = -50(420.25) + 42,102.50 - 20,700
f(20.50) = -21,012.50 + 42,102.50 - 20,700
f(20.50) = $18589.00
Therefore, the maximum profit is $18,589.00.
To find the price(s) that would enable the company to break even, set the profit function equal to 0:
-50p^2 + 2050p - 20,700 = 0
Solve for p using the quadratic formula:
p = (-2050 ± √(2050^2 - 4(-50)(-20,700))) / 2(-50)
p = (-2050 ± √(4202500 - 4140000)) / -100
p = (-2050 ± √622500) / -100
p = (-2050 ± 789.34) / -100
This gives us two possible prices that would enable the company to break even:
p1 = (-2050 + 789.34) / -100
p1 = 12.606
p2 = (-2050 - 789.34) / -100
p2 = 28.060
Therefore, the company would break even if they were to price the shirts at $12.61 and $28.06 per shirt.