A toy manufacturer's cost for producing q units of a game is given by
C(q) = 1480 + 3.8q + 0.0006q2. If the demand for the game is given by
p = 8.6 − 1/440q
how many games should be produced to maximize profit? (Round your answer to the nearest integer.)
revenue = price * demand
profit = revenue - cost
Hmmm. no indication of the selling price, unless that's p. If so, then revenue
r(q) = q*p(q) - c(q) = q(8.6-1/440 q) - (1480 + 3.8q + 0.0006q^2)
= -0.00287q^2 + 3.8q - 1480
For maximum revenue, now just find the vertex of the parabola. (where dr/dq = 0)
To maximize profit, we need to find the quantity of games that will yield the highest profit value.
To do this, we need to find the profit function, which is the difference between the revenue and the cost function.
The revenue function is given by R(q) = p * q, where p is the price and q is the quantity of games produced. We are given the demand function p = 8.6 - 1/440q, so we can substitute this into the revenue function:
R(q) = (8.6 - 1/440q) * q
Now, we have the cost function C(q) = 1480 + 3.8q + 0.0006q^2.
The profit function is calculated by subtracting the cost function from the revenue function:
Profit(q) = R(q) - C(q)
We can substitute the expressions for R(q) and C(q) into the profit function:
Profit(q) = ((8.6 - 1/440q) * q) - (1480 + 3.8q + 0.0006q^2)
Simplifying the expression gives:
Profit(q) = 8.6q - (1/440)q^2 - 1480 - 3.8q - 0.0006q^2
To find the quantity that maximizes profit, we need to find the critical points of the profit function. We can do this by taking the derivative of the profit function with respect to q, and setting it equal to zero.
Profit'(q) = 8.6 - (2/440)q - 3.8 - (0.0012)q
Setting the derivative equal to zero gives the following equation:
8.6 - (2/440)q - 3.8 - (0.0012)q = 0
Simplifying the equation gives:
4.8 - (2/440)q - (0.0012)q = 0
Now, we can solve this equation for q:
4.8 - (2/440)q - (0.0012)q = 0
4.8 = (2/440 + 0.0012)q
q = 4.8 / (2/440 + 0.0012)
Evaluate this expression to find the value of q that maximizes profit.