Problem 38 - College Algebra and Calculus: An Applied Approach

Question

A retail outlet sells two types of riding lawn mowers, the prices of which are \(p_{1}\) and \(p_{2}\). Find \(p_{1}\) and \(p_{2}\) so as to maximize total revenue, where \(R=515 p_{1}+805 p_{2}+1.5 p_{1} p_{2}-1.5 p_{1}^{2}-p_{2}^{2}\)

Step-by-Step Solution

Answer

To maximize total revenue, solve for \(p_{1}\) and \(p_{2}\) from the system of equations obtained from setting the first derivatives of the revenue function with respect to \(p_{1}\) and \(p_{2}\) to zero. Also, confirm that these solutions are a maximum by using the second derivative test. Exact values for \(p_{1}\) and \(p_{2}\) will depend on the solution to this system.

Step 1: Differentiate the revenue function with respect to \(p_{1}\)

In order to find the value of \(p_{1}\) at which revenue is maximized, differentiate \(R\) with respect to \(p_{1}\) and set the derivative equal to zero. The derivative is \(\frac{\partial R}{\partial p_{1}} = 515 + 1.5p_{2} - 3p_{1}\). Set this to zero to get: 515 + 1.5p_{2} - 3p_{1} = 0.

Step 2: Differentiate the revenue function with respect to \(p_{2}\)

Similarly, differentiate \(R\) with respect to \(p_{2}\) and set the derivative equal to zero. This derivative is \(\frac{\partial R}{\partial p_{2}} = 805 + 1.5p_{1} - 2p_{2}\). Set this to zero to get: 805 + 1.5p_{1} - 2p_{2} = 0.

Step 3: Solve the system of equations

We now have a system of two equations: 515 + 1.5p_{2} - 3p_{1} = 0 and 805 + 1.5p_{1} - 2p_{2} = 0. To find \(p_{1}\) and \(p_{2}\), we must solve this system of equations. The solution of this system will give the values for \(p_{1}\) and \(p_{2}\) that provide the maximum revenue.

Step 4: Make sure the solutions are maxima

Finally, we have to ensure that the solutions found actually correspond to a maximum of the revenue function. To do this, we can utilize the second derivative test. If the second derivative is less than zero for both \(p_{1}\) and \(p_{2}\), then the solutions are indeed a maximum.

Key Concepts

Differential Calculus

Differential Calculus is essential when you're trying to find out how to make more money from your products. Picture this: you're selling two different types of lawn mowers, and you want to set their prices so that you make the most profit possible. The total revenue depends on the prices, and we express it with a formula. This is where differential calculus comes in handy.

• First, by taking the derivative of the revenue function with respect to the prices \(p_1\) and \(p_2\), we can figure out how revenue changes as prices change.
• This involves applying the rules of differentiation, which are fundamental tools in calculus.

By calculating the derivative, we pinpoint where the revenue stops increasing and starts decreasing. This is key because that's the sweet spot where your revenue is maximized. So, differential calculus helps you make smart decisions about pricing.

Systems of Equations

When you have two different pricing strategies to look at, you end up with two equations. These equations come from setting each derivative found in the previous step to zero.

• The first equation is for \(p_1\): \(515 + 1.5p_2 - 3p_1 = 0\).
• The second equation is for \(p_2\): \(805 + 1.5p_1 - 2p_2 = 0\).

These are called a system of equations. To solve it, you're searching for values of \(p_1\) and \(p_2\) that will work with both equations simultaneously.
This might seem tricky, but it's manageable with methods such as substitution or elimination—powerful tools in algebra. Solving this system of equations gives you the price points that will bring in the maximum revenue.

Second Derivative Test

After finding potential solutions to our price problem, we need to confirm they actually maximize revenue. This is where the Second Derivative Test is incredibly useful.
To complete this test, you find the second derivatives of the revenue function with respect to both \(p_1\) and \(p_2\). If these second derivatives are negative, it means the price points found are indeed at a maximum.

• Calculate the second derivatives for both \(p_1\) and \(p_2\).
• Solve these to ensure the results are less than zero.

Simply put, the Second Derivative Test tells us whether the point found through the first derivative is a peak in the graph of the revenue function. This ensures you're not just finding any point, but the one that truly maximizes revenue.