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Chapter 12: Problem 76

If total demand \((Q)\) is specified by \(Q=-a P+b\), where \(P\) is unit price and \(a\) and \(b\) are positive parameters, then total revenue is maximized for this firm when it charges \(P\) equal to: (A) \(b / 2 a\) (B) \(b / 4 a\) (C) \(a / b\) (D) \(a / 2 b\) (E) \(-b / 2 a\)

Short Answer

Expert verified
The total revenue is maximized for this firm when it charges P equal to \(\frac{b}{2a}\).

Step by step solution

01

Express the total revenue (TR) function with respect to P

The total revenue is given by the product of the quantity demanded (Q) and the price (P). We have the demand equation as Q = -aP + b. To find the TR function, we'll multiply Q by P: TR = PQ Now, substitute the demand equation in the TR equation: TR = P(-aP + b)
02

Differentiate the TR function with respect to P

Now, we will differentiate the TR function with respect to P to find the critical points: \( \frac{d(TR)}{dP} = \frac{d(P(-aP + b))}{dP} \) Now, apply the product rule, \[ \frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \] where u = P and v = -aP + b: \( \frac{d(TR)}{dP} = P \frac{d(-aP + b)}{dP} + (-aP + b) \frac{dP}{dP} \)
03

Simplify the derivative and set it to zero

Now, we need to simplify the derivative and set it to zero to find the critical points: \( \frac{d(TR)}{dP} = P(-a) + (-aP + b)(1) \) Simplifying further: \( -aP - aP + b = 0 \) Now, set the derivative equal to zero: \( -2aP + b = 0 \)
04

Solve for P

Now, we'll solve for P, which represents the price that maximizes the total revenue: \( -2aP + b = 0 \) Add 2aP to both sides: \( b = 2aP \) Now, divide both sides by 2a: \( P = \frac{b}{2a} \) Comparing with the given options, this corresponds to option (A) \(b / 2 a\). Therefore, the total revenue is maximized for this firm when it charges P equal to \(\frac{b}{2a}\).

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Key Concepts

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

Understanding the Demand Function
The demand function is a key component in economics and it illustrates the relationship between the quantity of a product consumers are willing to buy and the price of that product. In this example, the demand function is given by the equation: \[ Q = -aP + b \] where:
  • \( Q \) represents the quantity demanded
  • \( P \) is the price of the product
  • \( a \) and \( b \) are parameters of the demand function, and they are both positive numbers.
The parameter \( a \) shows how much the quantity demanded changes with a change in price, it signifies the slope of the demand curve, and is often referred to as the price elasticity of demand. The parameter \( b \) represents the quantity demanded when the price is zero, acting much like the y-intercept of the demand curve.A negative sign in front of \( a \) highlights the inverse relationship between price and demand, indicating that as prices increase, demand decreases, which is a typical behavior in many markets. Understanding this relationship is essential for businesses as they strategize on pricing to maximize their revenue.
Revenue Function and How It Works
Once you understand the demand function, the next step is to analyze the revenue function. The revenue function represents how much money a company earns from selling its products at different price points. The total revenue (TR) function is derived by multiplying the quantity demanded \( Q \) by the price \( P \), resulting in:\[ TR = P(-aP + b) \]Expanding this, we get:\[ TR = -aP^2 + bP \]In this equation:
  • \( TR \) represents the total revenue
  • \(-aP^2\) indicates that revenue is affected by the square of the price, showing a quadratic relationship
  • \(bP\) represents the linear relationship of price to revenue
The quadratic form of this equation tells us that revenue reaches a peak at a certain price level, beyond which any increase in price actually causes a decrease in total revenue. This peak is precisely where businesses need to set their prices in order to maximize revenue.
Critical Points: Finding the Revenue Maximizing Price
To find the price that maximizes total revenue, we need to determine the critical points of the revenue function. These points occur where the derivative of the revenue function equals zero, helping identify where the maximum or minimum values occur. For our revenue function \( TR = -aP^2 + bP \), we calculate the derivative with respect to \( P \):\[ \frac{d(TR)}{dP} = -2aP + b \]Setting this derivative equal to zero solves for \( P \):\[ -2aP + b = 0 \]Rearranging gives us:\[ 2aP = b \]And finally, solving for \( P \):\[ P = \frac{b}{2a} \]This result means that the price to charge for maximizing total revenue is \( \frac{b}{2a} \). Understanding critical points in revenue functions is crucial for companies seeking to optimize profits, as it informs pricing strategies that align with market demand and economic principles.

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Most popular questions from this chapter

Sketch the graph of the given function, indicating (a) \(x\) - and \(y\) -intercepts, (b) extrema, (c) points of inflection, \((d)\) behavior near points where the function is not defined, and (e) behavior at infinity. Where indicated, technology should be used to approximate the intercepts, coordinates of extrema, and/or points of inflection to one decimal place. Check your sketch using technology. \(f(x)=-x^{2}-2 x-1\)

Given that the demand \(q\) is a differentiable function of the unit price \(p\), show that the revenue \(R=p q\) has a stationary point when $$q+p \frac{d q}{d p}=0$$ Deduce that the stationary points of \(R\) are the same as the points of unit price elasticity of demand. (Ordinarily, there is only one such stationary point, corresponding to the absolute maximum of \(R\).) HINT [Differentiate \(R\) with respect to \(p\).]

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Daily oil production in Mexico and daily U.S. oil imports from Mexico during \(2005-2009\) can be approximated by $$\begin{array}{ll}P(t)=3.9-0.10 t \text { million barrels } & (5 \leq t \leq 9) \\ I(t)=2.1-0.11 t \text { million barrels } & (5 \leq t \leq 9) \end{array}$$ where \(t\) is time in years since the start of \(2000 .^{41}\) Graph the function \(I(t) / P(t)\) and its derivative. Is the graph of \(I(t) / P(t)\) concave up or concave down? The concavity of \(I(t) / P(t)\) tells you that: (A) The percentage of oil produced in Mexico that was exported to the United States was decreasing. (B) The percentage of oil produced in Mexico that was not exported to the United States was increasing. (C) The percentage of oil produced in Mexico that was exported to the United States was decreasing at a slower rate. (D) The percentage of oil produced in Mexico that was exported to the United States was decreasing at a faster rate.
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