Problem 24
Question
Total sales, \(S,\) of Sea Change, Inc., 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 $$ M+L=70 $$
Step-by-Step Solution
Verified Answer
The maximum sales, subject to the constraint, are 612.5 when \(L = 17.5\) and \(M = 52.5\).
1Step 1: Define the Problem
We are given a function for total sales, \(S(L, M) = M L - L^2\), and a budget constraint \(M + L = 70\). Our task is to find the values of \(M\) and \(L\) that maximize \(S\) while satisfying the constraint.
2Step 2: Express M in Terms of L
Using the budget constraint \(M + L = 70\), express \(M\) in terms of \(L\):\[ M = 70 - L \]
3Step 3: Substitute M in the Sales Function
Substitute \(M = 70 - L\) into the sales function:\[ S(L, 70 - L) = (70 - L) L - L^2 \] Simplify it to:\[ S(L) = 70L - L^2 - L^2 = 70L - 2L^2 \]
4Step 4: Find the Derivative
Find the derivative of \(S(L)\) with respect to \(L\) to locate the critical points:\[ S'(L) = 70 - 4L \]
5Step 5: Set the Derivative to Zero
Set the derivative equal to zero and solve for \(L\):\[ 70 - 4L = 0 \ 4L = 70 \ L = \frac{70}{4} = 17.5 \]
6Step 6: Calculate M Using L
Substitute \(L = 17.5\) back into the expression for \(M\):\[ M = 70 - 17.5 = 52.5 \]
7Step 7: Calculate Maximum Sales
Substitute \(L = 17.5\) and \(M = 52.5\) into the sales function to find maximum sales:\[ S(17.5, 52.5) = 52.5 \times 17.5 - (17.5)^2 \]Calculate:\[ S(17.5, 52.5) = 918.75 - 306.25 = 612.5 \]
8Step 8: Verify Conditions
Check the second derivative test for maximum. Compute the second derivative: \[ S''(L) = -4 \]Since \(S''(L) < 0\), the function \(S\) has a local maximum at \(L = 17.5\).
Key Concepts
Budget ConstraintCritical PointsSecond Derivative Test
Budget Constraint
A budget constraint in optimization problems acts like a boundary for how resources can be allocated. Here, it is expressed by the equation \( M + L = 70 \). This means the sum of the costs of materials \( M \) and labor \( L \) must equal 70.
The budget constraint is crucial because it limits the possible values of \( M \) and \( L \) you can work with. Think of it like a fixed amount of money you can spend to maximize your profit or sales. Without this constraint, you could simply choose large values for \( M \) and \( L \) to push the sales up indefinitely. But in real life, such unrestricted spending is not possible.
For solving the problem at hand, we expressed \( M \) in terms of \( L \) by rearranging this equation to the formula \( M = 70 - L \). This substitution is key, as it simplifies the two-variable equation into a single-variable equation in terms of \( L \) so you can use calculus to find extremities in the function.
The budget constraint is crucial because it limits the possible values of \( M \) and \( L \) you can work with. Think of it like a fixed amount of money you can spend to maximize your profit or sales. Without this constraint, you could simply choose large values for \( M \) and \( L \) to push the sales up indefinitely. But in real life, such unrestricted spending is not possible.
For solving the problem at hand, we expressed \( M \) in terms of \( L \) by rearranging this equation to the formula \( M = 70 - L \). This substitution is key, as it simplifies the two-variable equation into a single-variable equation in terms of \( L \) so you can use calculus to find extremities in the function.
Critical Points
Critical points in calculus are where the derivative of a function is zero or undefined. These points can indicate places where you have a local maximum, local minimum, or saddle point. In the exercise, the derivative \( S'(L) \) is given as \( 70 - 4L \).
To find the critical points, we set this derivative to zero:
This value represents a potential extremum for the cost of labor that, when combined with the cost of materials \( M = 70 - L \), will maximize total sales. Be careful, though—finding a critical point is not the same as assuring it's a maximum or minimum. Once a critical point is found, it's important to apply further tests like the second derivative test to ascertain the nature of the extremum.
To find the critical points, we set this derivative to zero:
- \( 70 - 4L = 0 \)
This value represents a potential extremum for the cost of labor that, when combined with the cost of materials \( M = 70 - L \), will maximize total sales. Be careful, though—finding a critical point is not the same as assuring it's a maximum or minimum. Once a critical point is found, it's important to apply further tests like the second derivative test to ascertain the nature of the extremum.
Second Derivative Test
The second derivative test helps determine whether a critical point is a local maximum, local minimum, or neither. After finding the critical point at \( L = 17.5 \), we use the second derivative \( S''(L) \) to test the concavity of the function at this point.
For the given function, the second derivative is \( S''(L) = -4 \). The second derivative being negative indicates that the function is concave down at \( L = 17.5 \). In simple terms, this shape suggests that the hill of our sales function is being maximized at this spot.
For the given function, the second derivative is \( S''(L) = -4 \). The second derivative being negative indicates that the function is concave down at \( L = 17.5 \). In simple terms, this shape suggests that the hill of our sales function is being maximized at this spot.
- If \( S''(L) > 0 \), the function is concave up, representing a local minimum.
- If \( S''(L) < 0 \), like in this case, the function is concave down, pointing to a local maximum.
- If \( S''(L) = 0 \), the test is inconclusive, and further analysis would be needed.
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