The cost function for a product is C(q)=q3−60q2+1200q+1160 for 0≤q≤50 and a price per unit of $599.
a) What production level maximizes profit?
q =
b) What is the total cost at this production level?
cost = $
c) What is the total revenue at this production level?
revenue = $
d) What is the total profit at this production level?
profit = $
Thanks in advance for anyones help!
To find the production level that maximizes profit, we need to differentiate the profit function with respect to quantity (q) and set it equal to zero. Here's how you can solve each part of the problem to find the answers:
a) To find the production level that maximizes profit, we need to find the quantity (q) that corresponds to the maximum point of the profit function. The profit function can be calculated as follows:
Profit(q) = Revenue(q) - Cost(q)
The revenue function can be calculated by multiplying the quantity (q) by the price per unit ($599):
Revenue(q) = q * 599
The cost function is already given:
Cost(q) = q^3 - 60q^2 + 1200q + 1160
Now, we can substitute the revenue and cost functions into the profit function:
Profit(q) = q * 599 - (q^3 - 60q^2 + 1200q + 1160)
b) To find the total cost at the production level that maximizes profit, substitute the value of q into the cost function:
Total Cost = Cost(q)
c) To find the total revenue at the production level that maximizes profit, substitute the value of q into the revenue function:
Total Revenue = Revenue(q)
d) To find the total profit at the production level that maximizes profit, subtract the total cost from the total revenue:
Total Profit = Total Revenue - Total Cost
Let's solve these calculations step-by-step:
a) To find the production level that maximizes profit, we need to differentiate the profit function with respect to quantity (q) and set it equal to zero.
Profit(q) = q * 599 - (q^3 - 60q^2 + 1200q + 1160)
Differentiate the profit function:
Profit'(q) = 599 - (3q^2 - 120q + 1200)
Set the first derivative equal to zero and solve for q:
599 - (3q^2 - 120q + 1200) = 0
3q^2 - 120q + 1200 = 599
3q^2 - 120q + 601 = 0
Using the quadratic formula: q = (-b ± √(b^2 - 4ac)) / (2a)
a = 3, b = -120, c = 601
Solving the quadratic equation, we obtain two values for q. However, since the problem specifies the range of 0 ≤ q ≤ 50, we only consider the solution within this range.
b) Now that we have the production level (q) that maximizes profit, we can find the total cost at this production level. Substitute the obtained value of q into the cost function:
Total Cost = Cost(q)
Plug in the value of q into the cost function: Total Cost = q^3 - 60q^2 + 1200q + 1160
c) To find the total revenue at the production level that maximizes profit, substitute the value of q into the revenue function:
Total Revenue = Revenue(q)
Plug in the value of q into the revenue function: Total Revenue = q * 599
d) Calculate the total profit by subtracting the total cost from the total revenue:
Total Profit = Total Revenue - Total Cost
These steps will help you find the answers to the given questions.