Using the revenue function R(q) = 510q and the cost function C(q) = 9400+3q2. Round your answers to the nearest whole number.

a) At what quantity is profit maximized?
q =

b) What is the total profit at this production level?
profit = $

profit P(q) = R(q)-C(q), so

P(q) = 510q - (9400+3q^2)
dP/dq = 510 - 6p

P is maximized when dP/dq = 0, so ...

To find the quantity at which profit is maximized, we need to calculate the profit function, which is the difference between the revenue function and the cost function.

The profit function P(q) for a given quantity q is given by:
P(q) = R(q) - C(q)

Given the revenue function R(q) = 510q and the cost function C(q) = 9400 + 3q^2, we can substitute these into the profit function:

P(q) = 510q - (9400 + 3q^2)

Simplifying the profit function:

P(q) = 510q - 9400 - 3q^2

To find the quantity at which profit is maximized, we need to find the maximum value of the profit function. We can do this by finding the derivative of the profit function with respect to q and setting it equal to zero:

P'(q) = 510 - 6q

Setting P'(q) = 0:

510 - 6q = 0

Solving for q:

6q = 510
q = 510/6
q ≈ 85

Therefore, the quantity at which profit is maximized is approximately 85 (rounded to the nearest whole number).

To find the total profit at this production level, we can substitute the quantity q = 85 into the profit function P(q):

P(85) = 510(85) - 9400 - 3(85)^2

Calculating the profit:

P(85) ≈ 43350 - 9400 - 3(7225)
≈ 43350 - 9400 - 21675
≈ 12275

Therefore, the total profit at this production level is approximately $12,275 (rounded to the nearest whole dollar).