Business Functions: Cost, Revenue and Profit Functions PDF

Okay, here is the markdown format of the provided document: ## Business Functions ### Cost Function $C = \text{(fixed costs)} + \text{(variable cost)}$ $C = a + bx$ ### Price - Demand Function $p = m - nx$ where: $x \text{ is the number of items that can be sold at } $p \text{ per item}$ ### Rev...

Okay, here is the markdown format of the provided document: ## Business Functions ### Cost Function $C = \text{(fixed costs)} + \text{(variable cost)}$ $C = a + bx$ ### Price - Demand Function $p = m - nx$ where: $x \text{ is the number of items that can be sold at } $p \text{ per item}$ ### Revenue Function $R = \text{(number of items sold)} * \text{(price per item)}$ $R = xp$ $R = x(m - nx)$ ### Profit Function $P = R - C$ $P = x(m - nx) - (a + bx)$ ## Example 1 A company manufactures gaming headsets. Its marketing research department has determined that the data is modeled by the price-demand function $p(x) = 3,000 - 50x$, where $1 \leq x \leq 40$, (x is in thousands). **(a)** Determine the company's revenue function and state its domain. **(b)** Find the revenue generated when 30 headsets are sold. ### Solution: **(a)** $R(x) = x * p(x)$ $R(x) = x(3000 - 50x)$ $R(x) = 3000x - 50x^2$ The domain of this function is the same as the domain of the price-demand function, which is $1 \leq x \leq 40$ (in thousands). **(b)** $R(30) = 3000(30) - 50(30)^2$ $R(30) = 90,000 - 50(900)$ $R(30) = 45,000$ Revenue = $45,000,000 ## Example 2 The financial department for the company in example 1 has established the following cost function for producing and selling x thousand headsets: $C(x) = 5,000 + 400x$ (x is in thousand dollars). **(a)** Write a profit function for producing and selling x thousand headsets, and indicate the domain of this function. **(b)** Determine the number of headsets that should be sold to achieve the maximum profit. **(c)** State the maximum profit. **(d)** How many headsets should be sold for the company to break even? ### Solution: **(a)** $P(x) = R(x) - C(x)$ $P(x) = (3000x - 50x^2) - (5000 + 400x)$ $P(x) = 3000x - 50x^2 - 5000 - 400x$ $P(x) = -50x^2 + 2600x - 5000$ The domain of this function is the same as the domain of the original price-demand function, $1 \leq x \leq 40$ (in thousands.) **(b)** $P(x) = -50x^2 + 2600x - 5000$ -----> quadratic equation $a = -50$, $b = 2600$, $c = -5000$ Finding vertex: $h = \frac{-b}{2a}$, $k = f(h)$ $h = \frac{-2600}{2(-50)}$ $h = 26$ 26,000 headsets should be sold to achieve their maximum profit **(c)** Maximum profit: $f(h) = k$ $f(h) = -50(26)^2 + 2600(26) - 5000$ $f(h) = -50(676) + 67600 - 5000$ $f(h) = -33800 + 67600 - 5000$ $f(h) = 28800 \text{ (thousand dollars)}$ i.e. $28,800,000$ So the maximum profit is $28,800,000 and it is achieved when 26,000 headsets are sold. **(d**) For Break-Even, $P(x) = 0$ $-50x^2 + 2600x - 5000 = 0$ $-5x^2 + 260x - 500 = 0$ $a = -5, b = 260, c = -500$ $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $x = \frac{-(260) \pm \sqrt{(260)^2 - 4(-5)(-500)}}{2(-5)}$ $x = \frac{-260 \pm \sqrt{67600 - 10000}}{-10}$ $x = \frac{-260 \pm \sqrt{57600}}{-10}$ $x = \frac{-260 \pm 240}{-10}$ $x = \frac{-260 + 240}{-10}$ and $x = \frac{-260 - 240}{-10}$ $x = 2$ and $x = 50$ i.e. 2,000 units i.e. 50,000 units The image at the end contains a graph. It’s a parabola that opens downward. The vertex is labelled with (h, k) and is at the point (26, 28,800). The parabola intersects the x axis at x=2 and x=50. The x axis label is (thousands). The y axis label is P (thousands) and has a horizontal line at 28,800.