A producer of computer graphics software finds that selling price p of its software is related to the number of x copies of its software sold annually by the demand equation, x = 10000 - 200p, while its total cost in producing and marketing these x copies is given by the function C(x)= 50000 + 5x. Find the price p for which profits will be a maximum.
Find the maximum profit earned by selling at this price.
To find the price (p) for which profits will be maximum, we need to determine the profit function and then find the value of p that maximizes it.
The profit (P) is given by the equation: P = Revenue - Cost.
The revenue (R) can be found by multiplying the selling price (p) by the number of copies sold (x): R = p * x.
The cost (C) function is already given as C(x) = 50000 + 5x.
Substituting the demand equation into the revenue equation, we have: R = p * (10000 - 200p).
Now, we can express the profit function (P) in terms of p:
P = R - C
= p * (10000 - 200p) - (50000 + 5x)
= p * (10000 - 200p) - (50000 + 5(10000 - 200p)) [Substituting x = 10000 - 200p]
Simplifying further, we get:
P = 10000p - 200p^2 - 50000 - 50000 + 1000p
= -200p^2 + 11000p - 100000
To find the price p for which profits will be a maximum, we need to find the critical points of this profit function. We can do this by taking the derivative of P with respect to p and setting it equal to 0.
Let's differentiate P with respect to p:
P' = -400p + 11000
Setting P' = 0, we have:
-400p + 11000 = 0
-400p = -11000
p = (-11000) / (-400)
p = 27.5
Now, let's substitute the value of p back into the profit function to find the maximum profit:
P = -200p^2 + 11000p - 100000
P = -200(27.5)^2 + 11000(27.5) - 100000
P ≈ $406,250
Therefore, the price (p) for which profits will be a maximum is $27.5, and the maximum profit earned by selling at this price is approximately $406,250.