Raggs, Ltd. a clothing firm, determines that in order to sell $x$ suits, the price per suit must be $p=140-0.75 x$. It also determines that the total cost of producing $x$ suits is given by $C(x)=2500+0.5 x^{2}$.
What is the maximum profit?

Step 1 ：Let's denote the number of suits sold as \(x\), the price per suit as \(p\), and the total cost of producing \(x\) suits as \(C(x)\). Raggs, Ltd. has determined that \(p = 140 - 0.75x\) and \(C(x) = 2500 + 0.5x^2\).

Step 2 ：The profit is given by the difference between the total revenue and the total cost. The total revenue is given by the price per suit times the number of suits sold, which is \(p \cdot x = (140 - 0.75x) \cdot x\). The total cost is given by \(C(x) = 2500 + 0.5x^2\). Therefore, the profit function is given by \(P(x) = (140 - 0.75x) \cdot x - (2500 + 0.5x^2)\).

Step 3 ：To find the maximum profit, we need to find the maximum of the profit function. This can be done by finding the derivative of the profit function and setting it equal to zero, then solving for \(x\). The derivative of the profit function is \(P'(x) = 140 - 2.5x\).

Step 4 ：Solving \(P'(x) = 0\) gives us the critical points of the profit function. In this case, the critical point is \(x = 56\).

Step 5 ：Substituting \(x = 56\) back into the profit function gives us the maximum profit, which is \(P(56) = 1420\).

Step 6 ：Final Answer: The maximum profit is \(\boxed{1420}\).