A profit function is derived from the production cost and revenue function for a given item. The monthly profit function for a certain item is given by P(x)=−0.05x^2+400x−100,000, where P is in dollars and x is the number of units sold.
a. How many units must be sold on a monthly basis to maximize the profit?
b. Find the maximum profit
as you recall, the parabola
ax^2+bx+c has its vertex at
x = -b/2a
so, plug that in and use that to get part (b)
To answer these questions, we can use calculus. The concept of derivatives helps us find the maximum or minimum points of a function.
a. To find the number of units that must be sold to maximize the profit, we need to find the value of x that corresponds to the maximum value of the profit function.
To do this, we take the derivative of the profit function P(x) with respect to x, set it equal to zero, and solve for x.
P'(x) = -0.1x + 400 = 0
Solving this equation, we get:
-0.1x = -400
x = -400 / -0.1
x = 4000
Therefore, 4000 units must be sold on a monthly basis to maximize the profit.
b. To find the maximum profit, we substitute the value of x we found in part a (x = 4000) into the profit function P(x).
P(4000) = -0.05(4000)^2 + 400(4000) - 100,000
P(4000) = -0.05(16,000,000) + 1,600,000 - 100,000
P(4000) = -800,000 +1,600,000 - 100,000
P(4000) = 700,000
Thus, the maximum profit is $700,000.