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. show full work.
To find the maximum profit, we need to consider the profit function. Profit is given by the equation:
Profit = Total Revenue - Total Cost
Total Revenue is calculated by multiplying the selling price (p) by the quantity sold (q):
Total Revenue = p * q
Total Cost is given by the integral of the marginal cost function:
Total Cost = ∫(dc/dq) dq
Let's calculate the total cost:
∫(dc/dq) dq = ∫(800/q + 5) dq
= 800 ln|q| + 5q + C
Where C is a constant of integration.
Next, we substitute the demand equation (p = 400 - 50q) into the profit equation:
Profit = (400 - 50q) * q - (800 ln|q| + 5q + C)
To find the maximum profit, we need to maximize this function by taking its derivative with respect to q and setting it to zero.
d(Profit)/dq = 400q - 50q^2 - (800/q + 5) = 0
Let's solve this equation for q:
400q - 50q^2 - (800/q + 5) = 0
Multiplying through by q:
400q^2 - 50q^3 - 800 - 5q = 0
Rearranging the terms:
50q^3 - 400q^2 + 5q + 800 = 0
Unfortunately, this equation does not have a simple solution. We would need to use numerical methods or a graphing calculator to approximate the value of q that maximizes the profit.
Once we find the value of q that maximizes the profit, we can substitute it back into the demand equation (p = 400 - 50q) to find the selling price (p) for that quantity.
To find the maximum profit, we need to determine the quantity of jackets that maximizes the profit function. The profit function is given by:
Profit = Revenue - Cost
We can calculate revenue by multiplying the selling price (p) by the quantity of jackets sold (q):
Revenue = p * q
The cost function is given by the marginal cost (dc/dq) multiplied by the quantity of jackets (q):
Cost = (800/q + 5) * q
Now, we can substitute the given demand equation for p in the Revenue equation:
Revenue = (400 - 50q) * q
And substitute the cost function into the Cost equation:
Cost = (800/q + 5) * q
Now, let's calculate the total profit by subtracting the cost from the revenue:
Profit = Revenue - Cost
Profit = (400 - 50q) * q - (800/q + 5) * q
Simplify the equation:
Profit = (400q - 50q^2) - (800 + 5q)
Profit = 400q - 50q^2 - 800 - 5q
To find the maximum profit, we need to find the value of q that maximizes this profit function. We can do this by taking the derivative of the profit function with respect to q and setting it equal to zero:
d(Profit)/dq = 400 - 100q - 5
0 = 400 - 100q - 5
Solve for q:
400 - 100q - 5 = 0
-100q = -395
q = 3.95
Since the demand equation represents the number of jackets sold in thousands (q), we need to multiply the result by 1000:
q = 3.95 * 1000
q = 3950
Therefore, the number of jackets that must be sold to obtain the maximum profit is 3950.