C(q) = q3−60q2+1200q+760 for 0 ≤ q ≤ 50 and a price per unit of $551.

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 = $

did you check out the hints I gave on your previous post of this problem? If so, where do you get stuck?

The ticket problem is similar. How far do you get on that one?

Profit = Revenue - Cost

P = -q^3 + 60 q^2 +(-1200+551)q - 760

dp/dq = -3 q^2 +120q - 649
zero for max or min
q^2 - 40 q + 216.333... = 0

q = [ 40 +/-sqrt (1600-865)]/2
q = [40+/-27.1)/2
q = 34 or 6.5
to find out if max or min, take next derivative
d^2p/dq^2 = -6q +120
at q = 34
this is -84, so a maximum
at q = 6.5
this is +81 so a minimum
so for maximum profit we want q = 34
I think you can do the rest, check my arithmetic

To find the production level that maximizes profit, we need to find the value of q that corresponds to the maximum value of the profit function. The profit function is determined by subtracting the total cost function from the total revenue function.

a) To find the production level that maximizes profit, we need to find the critical points of the profit function. These critical points occur when the derivative of the profit function is equal to zero. Since the profit function is given by P(q) = R(q) - C(q), we need to find the derivative of R(q) and C(q) separately.

The total revenue function is given by R(q) = pq, where p is the price per unit. In this case, p = $551. Therefore, R(q) = 551q.

Differentiating R(q) with respect to q, we get dR/dq = 551.

The total cost function is given by C(q) = q^3 - 60q^2 + 1200q + 760.

To find the derivative of C(q), we differentiate each term individually. The derivative of q^3 is 3q^2, the derivative of -60q^2 is -120q, the derivative of 1200q is 1200, and the derivative of 760 is 0. Summing these derivatives, we get dC/dq = 3q^2 - 120q + 1200.

Now we can set dR/dq equal to dC/dq to find the critical point:

551 = 3q^2 - 120q + 1200.

Simplifying this equation, we have 3q^2 - 120q + 649 = 0.

To solve for q, we can either use factoring or the quadratic formula. Let's use the quadratic formula:

q = (-(-120) ± √((-120)^2 - 4(3)(649))) / (2(3)).

Simplifying further, we get:

q = (120 ± √(14400 - 7776)) / 6.

q = (120 ± √(6624)) / 6.

q = (120 ± 81.36) / 6.

This gives us two possible values for q: q ≈ 33.89 and q ≈ 20.11.

Since the domain for q is limited to 0 ≤ q ≤ 50, the value of q that maximizes profit is q ≈ 33.89.

b) To find the total cost at this production level, we substitute the value of q into the total cost function:

cost = C(33.89) = (33.89)^3 - 60(33.89)^2 + 1200(33.89) + 760.

Evaluating this expression, we get cost ≈ $36,944.

c) To find the total revenue at this production level, we substitute the value of q into the total revenue function:

revenue = R(33.89) = 551(33.89).

Evaluating this expression, we get revenue ≈ $18,654.

d) To find the total profit at this production level, we subtract the total cost from the total revenue:

profit = revenue - cost.

Using the values we found in parts b) and c), we have:

profit = $18,654 - $36,944.

Evaluating this expression, we get profit ≈ -$18,290.