Question

The price \(p\) (in dollars) and the quantity \(x\) sold of a certain product satisfy the demand equation $$ x=-6 p+600 $$ (a) Find a model that expresses the revenue \(R\) as a function of \(p .(\) Remember \(, R=x p .)\) (b) What is the domain of \(R ?\) Assume \(R\) is nonnegative. (c) What price \(p\) maximizes the revenue? (d) What is the maximum revenue? (e) How many units are sold at this price? (f) Graph \(\underline{R}\). (g) What price should the company charge to earn at least \(\$ 12,600\) in revenue?

Short answer

Revenue function: \(R(p) = -6p^2 + 600p\), Domain: \(0 \le p \le 100\), Maximum revenue at \(p=50\) is \$15,000 for 300 units. To earn at least \$12,600, charge \$30 or \$70.

Step-by-step solution

Express Revenue as a Function of p

First, use the demand equation to express the quantity sold, \(x\), in terms of price, \(p\). The given demand equation is \(x = -6p + 600\). Revenue \(R\) is given by \(R = x \, p\). Substitute \(x\) with \(-6p + 600\) in the revenue equation to get:\[R(p) = p(-6p + 600) = -6p^2 + 600p\]

Determine the Domain of R(p)

Revenue must be nonnegative, meaning \(R(p) \ge 0\). Since \(x\) represents quantity and must be nonnegative, \(-6p + 600 \ge 0\). Solving for \(p\) gives:\[p \le 100\]. Also, \(p \ge 0\) since price cannot be negative. Thus, the domain is: \[0 \le p \le 100\]

Find the Price that Maximizes Revenue

Revenue is a quadratic function \(R(p) = -6p^2 + 600p\) that opens downwards (since the coefficient of \(p^2\) is negative). The vertex gives the maximum revenue. For a quadratic function \(ax^2 + bx + c\), the vertex occurs at \(p = -\frac{b}{2a}\). Here, \(a = -6\) and \(b = 600\), so:\[p = -\frac{600}{2(-6)} = 50\]

Calculate the Maximum Revenue

Substitute \(p = 50\) into the revenue function to find the maximum revenue:\[R(50) = -6(50)^2 + 600(50) = -6(2500) + 30000 = -15000 + 30000 = 15000\]. Thus, the maximum revenue is \$15,000.

Determine the Quantity Sold at the Optimal Price

Substitute \(p=50\) into the demand equation to calculate \(x\):\[x = -6(50) + 600 = -300 + 600 = 300\]. So, 300 units are sold at the price that maximizes revenue.

Graph the Revenue Function

A graph of \(R(p) = -6p^2 + 600p\) is a downward-opening parabola with a vertex at \(p=50\) and \(R(50) = 15000\). The intercepts will be at \(p=0\) and \(p=100\).

Calculate the Price for at least \$12,600 in Revenue

Set \(-6p^2 + 600p = 12600\) and solve for \(p\):\[-6p^2 + 600p - 12600 = 0\]. This simplifies to \[p^2 - 100p + 2100 = 0\]. Using the quadratic formula \(p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we get:\[p = \frac{100 \pm \sqrt{10000 - 8400}}{2} = \frac{100 \pm 40}{2}\]. Thus, \(p = 70\) or \(p = 30\). Therefore, the company should charge either \$30 or \$70 to earn at least \$12,600.

Key concepts

Quadratic Functions

A quadratic function is a type of mathematical equation that can be written in the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. Quadratic functions form a parabola when graphed, which can either open upwards (if \(a > 0\)) or downwards (if \(a < 0\)). In the context of revenue optimization, quadratic functions can help model the relationship between different variables to understand how changes impact outcomes. The vertex of the parabola represents the maximum or minimum point, providing critical insights like the maximum revenue.

Revenue Function

The revenue function represents the total income a company receives from selling a certain number of goods or services at different prices. It is calculated by multiplying the quantity of goods sold \(x\) by the price per unit \(p\). Using the demand equation \(x = -6p + 600\), you can substitute \(x\) into the revenue formula, giving us \(R(p) = p (-6p + 600) = -6p^2 + 600p\). This equation is a quadratic function where the price \(p\) is the variable.

Demand Equation

The demand equation shows the relationship between the price of a product and the quantity sold. In this case, the demand equation is given as \(x = -6p + 600\). This linear equation indicates that for every increase in price, the quantity sold decreases by 6 units. By incorporating this into the revenue function, you can predict how changes in price affect the number of units sold and, therefore, the overall revenue.

Domain and Range

The domain and range of a function are critical to understanding its behavior. The domain of the revenue function \(R(p) = -6p^2 + 600p\) consists of all possible values of \(p\) that make sense in the given context. Here, the domain is \(0 \le p \le 100\) because prices below 0 or above 100 are not meaningful. The range of the function pertains to all the possible values of the revenue \(R\), which lies between 0 and its maximum value of 15,000, when considering the vertex of the parabola.