A company decides to begin making and selling computers. The price function is given as follows: $$ p=-50 x+2300 $$ where $$ x $$ is the number of computers that can be sold at a price of $$ p $$ dollars per unit. Additionally, the financial department has determined that the weekd) fixed cost of production will be 7000 dollars with an additional cost of 100 dollars per unit. (A) Find the revenue function in terms of $$ x $$. $$ R(x)=\square $$ (B) Use the financial department's estimates to determine the cost function in terms of $$ x $$. $$ C(x)=\square $$ (C) Find the profit function in terms of $$ x $$. $$ P(x)=\square $$ (D) Evaluate $$ P^{\prime}(x) $$ at $$ x=250 $$. $$ P^{\prime}(250)=\square $$

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(A) $$ R(x) = -50x^2 + 2300x $$ (B) $$ C(x) = 7000 + 100x $$ (C) $$ P(x) = -50x^2 + 2200x - 7000 $$ (D) $$ P'(250) = -22800 $$ To solve the problem, we will go through each part step by step. ### (A) Find the revenue function in terms of $$ x $$. The revenue function $$ R(x) $$ is given by the product of the price per unit $$ p $$ and the number of units sold $$ x $$. From the price function: $$ p = -50x + 2300 $$ The revenue function is: $$ R(x) = p \cdot x = (-50x + 2300) \cdot x $$ $$ R(x) = -50x^2 + 2300x $$ So, the revenue function is: $$ R(x) = -50x^2 + 2300x $$ ### (B) Determine the cost function in terms of $$ x $$. The cost function $$ C(x) $$ includes fixed costs and variable costs. The fixed cost is $7000, and the variable cost is $100 per unit produced. Thus, the cost function is: $$ C(x) = \text{Fixed Cost} + \text{Variable Cost} \cdot x $$ $$ C(x) = 7000 + 100x $$ ### (C) Find the profit function in terms of $$ x $$. The profit function $$ P(x) $$ is defined as the revenue function minus the cost function: $$ P(x) = R(x) - C(x) $$ Substituting the expressions we found for $$ R(x) $$ and $$ C(x) $$: $$ P(x) = (-50x^2 + 2300x) - (7000 + 100x) $$ $$ P(x) = -50x^2 + 2300x - 7000 - 100x $$ $$ P(x) = -50x^2 + 2200x - 7000 $$ ### (D) Evaluate $$ P^{\prime}(x) $$ at $$ x=250 $$. First, we need to find the derivative of the profit function $$ P(x) $$: $$ P(x) = -50x^2 + 2200x - 7000 $$ Taking the derivative: $$ P'(x) = -100x + 2200 $$ Now, we evaluate $$ P'(x) $$ at $$ x = 250 $$: $$ P'(250) = -100(250) + 2200 $$ $$ P'(250) = -25000 + 2200 $$ $$ P'(250) = -22800 $$ ### Summary of Results: (A) $$ R(x) = -50x^2 + 2300x $$ (B) $$ C(x) = 7000 + 100x $$ (C) $$ P(x) = -50x^2 + 2200x - 7000 $$ (D) $$ P'(250) = -22800 $$

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