Raggs, Ltd. a clothing firm, determines that in order to sell $x$ suits, the price per suit must be $p=120-0.75 x$. It also determines that the total cost of producing $x$ suits is given by $C(x)=2000+0.75 x^{2}$ a) Find the total revenue, $R(x)$. b) Find the total profit, $P(x)$. c) How many suits must the company produce and sell in order to maximize profit? d) What is the maximum profit? e) What price per suit must be charged in order to maximize profit?

Step 1 ：The total revenue, $R(x)$, is given by the price per suit, $p$, times the number of suits sold, $x$. So, $R(x) = p \cdot x = (120 - 0.75x) \cdot x = 120x - 0.75x^2$.

Step 2 ：The total profit, $P(x)$, is given by the total revenue, $R(x)$, minus the total cost, $C(x)$. So, $P(x) = R(x) - C(x) = (120x - 0.75x^2) - (2000 + 0.75x^2) = 120x - 2000$.

Step 3 ：To find the number of suits the company must produce and sell in order to maximize profit, we need to find the value of $x$ that maximizes $P(x)$. This occurs where the derivative of $P(x)$ is zero. So, we find the derivative of $P(x)$, set it equal to zero, and solve for $x$. The derivative of $P(x)$ is $P'(x) = 120 - 1.5x$. Setting this equal to zero gives $120 - 1.5x = 0$, which solves to $x = \frac{120}{1.5} = 80$.

Step 4 ：The maximum profit is $P(80) = 120 \cdot 80 - 2000 = 9600 - 2000 = 7600$.

Step 5 ：The price per suit that must be charged in order to maximize profit is $p = 120 - 0.75 \cdot 80 = 120 - 60 = 60$.