the cost function C(q) = q3−57q2+1083q+1010 for 0 ≤ q ≤ 50 and a price per unit of $534.
Round your answers to the nearest whole number.
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 = $
To find the production level that maximizes profit, we need to differentiate the cost function with respect to q and find where the derivative is equal to 0. Let's start by finding the derivative of the cost function.
C'(q) = 3q^2 - 114q + 1083
Now, set the derivative equal to 0 and solve for q.
3q^2 - 114q + 1083 = 0
We can solve this quadratic equation using the quadratic formula.
q = (-(-114) ± √((-114)^2 - 4(3)(1083))) / (2(3))
This simplifies to:
q = (114 ± √(13056)) / 6
q = (114 ± 114.28) / 6
Now, we need to check if these solutions fall within the range 0 ≤ q ≤ 50.
For q = (114 + 114.28) / 6 ≈ 37.38, this solution is within the range.
For q = (114 - 114.28) / 6 ≈ -0.047, this solution is not within the range.
So, the production level that maximizes profit is q ≈ 37.38.
Now, let's find the total cost at this production level by plugging q ≈ 37.38 into the cost function.
cost = (37.38)^3 - 57(37.38)^2 + 1083(37.38) + 1010 ≈ $43,598.00
Next, we need to calculate the total revenue at this production level. Recall that revenue is equal to the price per unit multiplied by the production level.
revenue = $534 * 37.38 ≈ $19,989.72
Finally, let's calculate the total profit by subtracting the total cost from the total revenue.
profit = $19,989.72 - $43,598.00 ≈ -$23,608.28
Therefore:
a) The production level that maximizes profit is q ≈ 37.38.
b) The total cost at this production level is approximately $43,598.
c) The total revenue at this production level is approximately $19,990.
d) The total profit at this production level is approximately -$23,608.
To find the production level that maximizes profit, we need to first find the derivative of the cost function and set it equal to zero. The derivative represents the rate at which the cost function is changing, and when it equals zero, it implies that the cost function is at a maximum or minimum.
a) To find the production level that maximizes profit, we need to differentiate the cost function C(q) with respect to q:
C'(q) = 3q^2 - 114q + 1083
Setting the derivative equal to zero, we get:
3q^2 - 114q + 1083 = 0
Now we can solve this quadratic equation to find the value of q that maximizes profit. We can use the quadratic formula:
q = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 3, b = -114, and c = 1083. Plugging in these values, we get:
q = (-(-114) ± √((-114)^2 - 4 * 3 * 1083)) / (2 * 3)
q = (114 ± √(12996 - 12996)) / 6
q = 114 / 6
q = 19
Therefore, the production level that maximizes profit is q = 19.
b) To find the total cost at this production level, we substitute q = 19 into the cost function C(q):
C(19) = (19)^3 - 57(19)^2 + 1083(19) + 1010
Calculating this expression, we get:
C(19) = 685 - 20283 + 20577 + 1010
C(19) = 9029
The total cost at this production level is $9029.
c) To find the total revenue at this production level, we multiply the production level q by the price per unit:
revenue = q * price per unit
revenue = 19 * $534
revenue = $10146
The total revenue at this production level is $10146.
d) To find the total profit at this production level, we subtract the total cost from the total revenue:
profit = revenue - cost
profit = $10146 - $9029
profit = $1117
The total profit at this production level is $1117.