The revenue from selling q items is R(q)=475q−q2, and the total cost is C(q)=175+6q. Write a function that gives the total profit earned, and find the quantity which maximizes the profit.

P = R-C

that will be a parabola. Find its vertex.

To find the total profit, we need to subtract the total cost from the revenue. The profit function, P(q), can be calculated as:

P(q) = R(q) - C(q)

Substituting the given revenue and cost functions into the profit function:

P(q) = (475q - q^2) - (175 + 6q)

Simplifying:

P(q) = 475q - q^2 - 175 - 6q

P(q) = -q^2 + 469q - 175

To find the quantity that maximizes the profit, we need to find the value of q that corresponds to the maximum value of the profit function. This can be done by finding the vertex (maximum point) of the quadratic equation. The x-coordinate of the vertex is given by:

q = -b / (2a)

In our case, the quadratic equation is -q^2 + 469q - 175, so a = -1 and b = 469. Plugging these values into the formula for the x-coordinate of the vertex:

q = -469 / (2*(-1))

q = -469 / (-2)

q = 234.5

Therefore, the quantity that maximizes the profit is 234.5 units.

To summarize:
1. The profit function is P(q) = -q^2 + 469q - 175.
2. The quantity that maximizes the profit is q = 234.5 units.