A manufacturing company sells high quality jackets through a chain of specialty shops. the demand equation for this jackets is
p = 400 − 50q
where p is the selling price ( in dollars per jacket) and q is the demand ( in thousands of jackets). If this company’s marginal cost function is given by
dc/dq =800/q + 5
show that there is a maximum profit, and determine the number of jackets that must be sold to obtain the maximum profit.
revenue = q*p(q) = 400q - 50q^2
So the profit is
P(q) = (400q - 50q^2) - c(q)
dP/dq = 400 - 100q - (800/q + 5)
Hmmm. dP/dq is never zero. Better check my math.
To determine the maximum profit, we need to find the quantity of jackets at which the marginal cost equals the marginal revenue. Then, we can calculate the profit at that quantity.
The marginal cost is given by the function dc/dq = 800/q + 5.
The marginal revenue is the derivative of the demand equation with respect to q, which is the negative of the slope of the demand equation.
So, the marginal revenue is given by the function -dp/dq = -(-50) = 50.
Setting the marginal cost equal to the marginal revenue, we have:
800/q + 5 = 50
Subtracting 5 from both sides:
800/q = 45
Multiplying both sides by q:
800 = 45q
Dividing both sides by 45:
q = 800/45 = 160/9 ≈ 17.78
Since the demand, q, represents thousands of jackets, we can multiply it by 1000 to find the actual quantity:
Q = 17.78 * 1000 ≈ 17778 jackets
Therefore, in order to obtain the maximum profit, the company must sell approximately 17,778 jackets.