Business Functions: Cost, Revenue, and Profit
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Questions and Answers

Explain how the price-demand function influences the revenue function. How does altering the price per item affect the total revenue?

The price-demand function dictates the price at which a certain quantity of items can be sold. This price directly multiplies by the number of items sold ($x$) to determine the revenue: $R = x * p(x)$. Altering the price per item (p) will change the quantity sold (x), subsequently affecting total revenue.

Describe the relationship between the cost function, revenue function, and profit function. How do changes in cost and revenue affect profit?

Profit is the difference between revenue and cost: $P = R - C$. An increase in cost, without a corresponding increase in revenue, will decrease profit, and vice versa. Similarly, an increase in revenue, without a corresponding increase in cost, will increase profit.

A company's cost function is given by $C(x) = 1000 + 20x$, and its revenue function is $R(x) = 50x$. Determine the number of items that must be sold for the company to break even.

To break even, $R(x) = C(x)$. Therefore, $50x = 1000 + 20x$. Solving for x: $30x = 1000$, so $x = rac{100}{3} \approx 33.33$. Since you can't sell fractions of items, the company needs to sell 34 items to break even.

The price-demand function for a product is given by $p(x) = 200 - 2x$. Find the revenue function and determine the quantity $x$ that maximizes revenue.

<p>The revenue function is $R(x) = x * p(x) = x(200 - 2x) = 200x - 2x^2$. To maximize revenue, find the vertex of the quadratic function. The $x$-value of the vertex is $x = -b / 2a = -200 / (2 * -2) = 50$. Therefore, the quantity that maximizes revenue is 50.</p> Signup and view all the answers

Explain the significance of finding the vertex of the profit function. What information does it provide, and how can it be used for decision-making?

<p>The vertex of the profit function represents the point at which profit is maximized (or minimized, if the parabola opens downward). The $x$-coordinate of the vertex indicates the quantity of items that should be produced/sold to achieve the maximum profit, and the $y$-coordinate represents the maximum profit value. This information helps in determining optimal production levels.</p> Signup and view all the answers

A company has a fixed cost of $5,000 and a variable cost of $50 per item. If the selling price per item is $100, write the cost, revenue, and profit functions. What is the profit if 200 items are sold?

<p>The cost function is $C(x) = 5000 + 50x$. The revenue function is $R(x) = 100x$. The profit function is $P(x) = R(x) - C(x) = 100x - (5000 + 50x) = 50x - 5000$. If 200 items are sold, the profit is $P(200) = 50(200) - 5000 = 10000 - 5000 = 5000$.</p> Signup and view all the answers

Suppose a company determines that its profit function is $P(x) = -2x^2 + 80x - 600$, where $x$ is the number of units sold. Determine the range of units the company needs to sell to make a profit.

<p>To find the range of units needed to make a profit, we need to find when $P(x) &gt; 0$. This requires finding the roots (or zeros) of the quadratic equation: $-2x^2 + 80x - 600 = 0$. Dividing by -2, we get: $x^2 - 40x + 300 = 0$. Factoring, we get $(x - 10)(x - 30) = 0$. Therefore, the roots are $x = 10$ and $x = 30$. The company needs to sell between 10 and 30 units to make a profit.</p> Signup and view all the answers

The price-demand function for a new gadget is modeled by $p = 50 - 0.5x$, where $x$ is the number of gadgets demanded. If the company's cost function is $C(x) = 10x + 500$, find the profit function and determine the number of gadgets that need to be sold to maximize profit.

<p>First, find the revenue function: $R(x) = x * p = x(50 - 0.5x) = 50x - 0.5x^2$. Then, the profit function is $P(x) = R(x) - C(x) = (50x - 0.5x^2) - (10x + 500) = -0.5x^2 + 40x - 500$. To maximize profit, find the vertex: $x = -b / 2a = -40 / (2 * -0.5) = 40$. Therefore, 40 gadgets need to be sold to maximize profit.</p> Signup and view all the answers

Flashcards

Cost Function (C)

Total cost to produce items; sum of fixed and variable costs. C = a + bx

Price-Demand Function (p)

Relates the price per item to the number of items that can be sold. p = m - nx

Revenue Function (R)

Total income from selling items. (number of items sold) * (price per item). R = xp

Profit Function (P)

The difference between revenue and cost. P = R - C

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How do you Determine the Revenue Function?

Substitute the demand equation into the revenue eauation. R(x) = x * p(x)

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How do you Determine the Domain of a Price-Demand Function?

Set the price-demand values as an inequality to find the domain of the function. (e.g. 1 ≤ x ≤ 40)

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How to determine the Profit function?

Substitute the Revenue(R(x)) and Cost(C(x)) functions into this equation: P(x) = R(x) - C(x)

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Maximize Profit

Find the x-value of the vertex using: h = -b / (2a).

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Study Notes

  • Study notes on business functions

Cost Function

  • C = fixed costs + variable cost
  • C = a + bx, where 'a' represents fixed costs and 'bx' represents variable costs.

Price-Demand Function

  • p = m - nx: 'x' is the number of items sold at a price of '$p' per item.

Revenue Function

  • R = (number of items sold) * (price per item)
  • R = xp or x(m - nx)

Profit Function

  • P = R - C
  • P = x(m - nx) - (a + bx)

Example 1: Gaming Headsets

  • A company's price-demand function for gaming headsets is p(x) = 3,000 – 50x
  • Here, 1 ≤ x ≤ 40, and 'x' is in thousands.
  • Revenue function: R(x) = x * p(x) = x(3000 - 50x) = 3000x - 50x²
  • The domain of the revenue function is the same as the price-demand function: 1 ≤ x ≤ 40 (in thousands).
  • When 30,000 headsets are sold (x=30): R(30) = 3000(30) - 50(30)² = $45,000,000

Example 2: Cost Function for Gaming Headsets

  • Cost function is C(x) = 5,000 + 400x (x is in thousands of dollars).
  • Profit function: P(x) = R(x) - C(x) = (3000x - 50x²) - (5000 + 400x) = -50x² + 2600x - 5000
  • The domain is the same as the price-demand function: 1 ≤ x ≤ 40 (in thousands).
  • Vertex: x = -b / 2a = - 2600 / (2 * -50) = 26 is the number of headsets (in thousands) to maximize profit.
  • Maximum profit: f(h) = -50(26)² + 2600(26) - 5000 = $28,800 thousand

Break-Even Point

  • Break-even point is where P(x) = 0
  • Solving -50x² + 2600x - 5000 = 0, simplifies to -5x² + 260x - 500=0, with coefficients a = -5, b = 260, c = -500
  • x = (-b ± √(b² - 4ac)) / (2a) = 2 and 50,000 units
  • Break-even points are at 2,000 units and 50,000 units

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Description

Explore cost, revenue, and profit functions in business. Understand fixed and variable costs, price-demand relationships, and profit calculation. Examples included relate to gaming headsets, and demonstrate practical applications.

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