A manufacturer of a product has a monthly cost for producing x items given by C(x) = 10+2x. The selling price function for this product is p(x) = 50-.01x. Find the maximum profit the company can expect monthly
the revenue function is price * quantity, so
R(x) = x(50-.01x)
profit is revenue less cost, so
P(x) = R(x)-C(x)
= x(50-.01x) - (10+2x)
= -0.01x^2 + 48x - 10
Now just set dP/dx=0 and evaluate P there.
Or, just use your Algebra I skills and find the vertex of the parabola.
profit = p(x) - C(x)
= 50-.01 x - 10 - 2x
= 40-2.01 x
max at x = 0
I bet you have a typo though.
Oh, go with what Steve said.
To find the maximum profit the company can expect monthly, we need to determine the quantity of items that will maximize the profit.
The profit function is given by P(x) = p(x) - C(x), where P(x) represents the profit.
Substituting the given selling price function p(x) = 50 - 0.01x and the cost function C(x) = 10 + 2x into the profit function, we have:
P(x) = (50 - 0.01x) - (10 + 2x)
= 50 - 0.01x - 10 - 2x
= 40 - 2.01x
To find the maximum profit, we need to find the value of x that maximizes P(x).
One approach is to graph the profit function P(x) and determine the highest point on the graph, which represents the maximum profit. However, since we have a simple linear function for P(x), we can use calculus to find the maximum.
To find the maximum, we take the derivative of P(x) with respect to x and set it equal to zero:
P'(x) = -2.01
Setting P'(x) equal to zero, we have:
-2.01 = 0
Since there are no solutions to this equation, it means that the profit function P(x) does not have a maximum. Instead, it is a linear function that decreases as x increases. Therefore, the maximum profit cannot be determined from this given data.
If you have additional information or constraints, such as a maximum production capacity or a minimum number of items to produce, please provide that information so we can further analyze and determine the maximum profit.