Problem 4 The price $p$ (in dollars) and... [FREE SOLUTION]

Chapter 4: Problem 4

The price $p$ (in dollars) and the quantity $x$ sold of a certain product satisfy the demand equation $$ x=-3 p+360 $$ (a) Find a model that expresses the revenue $R$ as a function of $p$. (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 $R$. (g) What price should the company charge to earn at least $\$ 9600$ in revenue?

Short Answer

Expert verified

The price that maximizes revenue is $\$60$, yielding a maximum revenue of $\$10800$ with 180 units sold. To achieve at least \$9600, set price to $\$40$ or $\$80$.

Step by step solution

Express Quantity in terms of Price

The demand equation is given by \[ x = -3p + 360 \].

Define Revenue Function

Revenue $R$ is defined as the product of price $p$ and quantity $x$. Therefore, we have: \[ R(p) = p \times x \]. Substituting the expression for $x$ from Step 1, we get: \[ R(p) = p(-3p + 360) \].

Simplify Revenue Function

Expand the equation to express revenue purely in terms of $p$: \[ R(p) = -3p^2 + 360p \]. This represents the revenue function.

Determine the Domain of Revenue

Revenue $R(p)$ must be nonnegative, \[-3p^2 + 360p \geq 0\]. Solve this inequality by factoring: \[ p(360 - 3p) \geq 0 \]. This requires both factors to be nonnegative. Thus, \[ 0 \leq p \leq 120 \]. This is the domain of $R(p)$.

Find Price that Maximizes Revenue

To maximize revenue, find the critical points of $R(p) = -3p^2 + 360p$. Take the derivative and set it to zero: \[ R'(p) = -6p + 360 = 0 \]. Solving the equation, we get: \[ p = 60 \].

Calculate Maximum Revenue

Substitute $ p = 60 $ back into the revenue function to get: \[ R(60) = -3(60)^2 + 360(60) \]. Calculation gives: \[ R(60) = 10800 \]. Thus, the maximum revenue is $\$10800$.

Determine Quantity Sold at Maximum Price

Find the quantity sold when $ p = 60 $ using the demand equation: \[ x = -3(60) + 360 = 180 \]. So, 180 units are sold.

Graph Revenue Function

Plot the quadratic function $R(p) = -3p^2 + 360p$ within the domain $0 \leq p \leq 120$. The graph will be a parabola opening downwards with vertex at $p = 60$.

Determine Price for Specific Revenue Target

To earn at least \$9600 in revenue, solve the inequality: \[-3p^2 + 360p \geq 9600 \]. Rearrange to form a quadratic equation: \[-3p^2 + 360p - 9600 = 0 \]. Use the quadratic formula to solve for $p$: \[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], with $a = -3$, $b = 360$, and $c = -9600$. Solving gives $ p = 40$ or $p = 80$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Revenue Function

In any business, the revenue function, denoted as $R(p)$, represents the total money earned from selling a product. It can be calculated by multiplying the price per unit ($p$) by the number of units sold ($x$). For this exercise, the demand equation $x = -3p + 360$ shows the relationship between price and quantity. Substituting $x$ into $R(p) = p \times x$ gives the revenue function: \ $R(p) = p(-3p + 360)$. This can be simplified to $R(p) = -3p^2 + 360p$, which is a quadratic equation. Understanding the revenue function helps determine how changes in price affect total revenue.

Demand Equation

The demand equation tells us how the quantity of a product sold ($x$) is related to its price ($p$). In this exercise, the demand equation is $x = -3p + 360$. This equation indicates that if the price increases, the quantity sold decreases, and vice versa. By analyzing the demand equation, businesses can set prices strategically to influence the quantity sold. Plugging this equation into the revenue formula helps in creating a model that shows the overall revenue based on different price points.

Quadratic Equation

A quadratic equation is of the form $ax^2 + bx + c = 0$. In the context of this revenue model, the revenue function $R(p) = -3p^2 + 360p$ is a quadratic equation. This equation forms a parabola when graphed. The parabola opens downward since the coefficient of $p^2$ is negative. The vertex of this parabola represents the maximum revenue point. Finding the vertex involves calculating the price that maximizes revenue and helps businesses understand the optimal selling price.

Domain of a Function

The domain of a function refers to all possible values that the input (in this case, price $p$) can take. For the revenue function $R(p) = -3p^2 + 360p$, the domain is determined by ensuring revenue ($R$) is non-negative. Factoring the inequality $-3p^2 + 360p \geq 0$ provides the domain $0 \leq p \leq 120$. This means the price must be within this range for the revenue to be valid. Understanding the domain helps in setting practical price limits.

Derivative

The derivative of a function, denoted as $f'(x)$, gives the rate at which the function's value changes with respect to its input. Finding the derivative of the revenue function $R(p) = -3p^2 + 360p$ gives $R'(p) = -6p + 360$. Setting this derivative to zero ($R'(p) = 0$) helps identify critical points where revenue could be maximized. Solving $-6p + 360 = 0$ yields $p = 60$. This tells us that the price at which revenue is maximized is $60. Thus, derivatives play a crucial role in optimization problems like finding maximum revenue.

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