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Chapter 6: Problem 21

The total sales, \(S\), of a oneproduct firm are given by \(S(L, M)=M L-L^{2}\) where \(M\) is the cost of materials and \(L\) is the cost of labor. Find the maximum value of this function subject to the budget constraint \(M+L=90\)

Short Answer

Expert verified
The maximum value of the sales function is 1012.5.

Step by step solution

01

- Define the objective function

The total sales function is given by \[ S(L, M) = M L - L^2 \]. This is our objective function that needs to be maximized.
02

- Define the constraint

The budget constraint is given by \[ M + L = 90 \].
03

- Express the constraint in terms of one variable

Express the cost of materials in terms of the cost of labor using the budget constraint:\[ M = 90 - L \].
04

- Substitute the constraint into the objective function

Substitute the value of \( M \) into the objective function:\[ S(L) = (90 - L) L - L^2 \].
05

- Simplify the objective function

Simplify the objective function:\[ S(L) = 90L - L^2 - L^2 = 90L - 2L^2 \].
06

- Differentiate the objective function

Differentiate the simplified objective function with respect to \( L \):\[ \frac{dS(L)}{dL} = 90 - 4L \].
07

- Find the critical points

Set the derivative equal to zero to find the critical points:\[ 90 - 4L = 0 \rightarrow 4L = 90 \rightarrow L = 22.5 \].
08

- Verify the maximum value using the second derivative

Find the second derivative of \( S(L) \):\[ \frac{d^2S(L)}{dL^2} = -4 \].Since the second derivative is negative, the function has a maximum at \( L = 22.5 \).
09

- Calculate the corresponding value of M

Using the constraint, calculate \( M \) when \( L = 22.5 \):\[ M = 90 - 22.5 = 67.5 \].
10

- Find the maximum value of the function

Substitute \( L = 22.5 \) and \( M = 67.5 \) back into the original objective function to find the maximum sales:\[ S(22.5, 67.5) = 67.5 \cdot 22.5 - (22.5)^2 = 1518.75 - 506.25 = 1012.5 \].

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Key Concepts

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

Objective Function
In optimization problems, the objective function is the main equation we are trying to maximize or minimize. For this example, our objective function is given as \[ S(L, M) = M L - L^{2} \].
This function represents the total sales of a one-product firm, where \( M \) is the cost of materials and \( L \) is the cost of labor.
The objective is to find the values of \( L \) and \( M \) that will maximize \( S(L, M) \). We use this function throughout the problem to determine the best possible sales for the given budget constraint.
Budget Constraint
A budget constraint restricts the possible values for our variables due to limited resources. Here, the budget constraint is given by \[ M + L = 90 \].
This means the combined cost of materials and labor must equal 90.
Constraints like this are common in real-world problems where resources are limited. By substituting this constraint into our objective function, we can simplify the problem to find optimal values for \( L \) and \( M \).
Specifically, we express \( M \) in terms of \( L \): \( M = 90 - L \). This substitution turns the problem into one that is easier to handle with a single variable.
Critical Points
Critical points of a function are the values of the variables that potentially provide the maximum or minimum values of the function.
To find these points, we first take the derivative of the simplified objective function:\( S(L) = 90L - 2L^2 \) with respect to \( L \): \( \frac{dS(L)}{dL} = 90 - 4L \).
We set the derivative equal to zero: \( 90 - 4L = 0 \), which gives us the critical point at \( L = 22.5 \).
This step is crucial because it tells us where the maximum or minimum of the function occurs.
Second Derivative Test
The second derivative test helps us verify whether a critical point is a maximum or minimum.
For our objective function \( S(L) = 90L - 2L^2 \), the second derivative is \( \frac{d^2S(L)}{dL^2} = -4 \).
A negative second derivative, as in this case, indicates a maximum point because the function is concave down at that critical point.
Hence, at \( L = 22.5 \), we have a maximum value for the function.
This final step confirms we have indeed found the best possible values for maximizing total sales given our budget constraint.

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Most popular questions from this chapter

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