Problem 23 - Calculus and its applications

Question

Total sales, \(S,\) of Cre-Tech 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 $$ M+L=90 $$

Step-by-Step Solution

Answer

The maximum sales, given the constraint, are 1012.5.

Step 1: Identify the Constraint Equation

The budget constraint given is \( M+L=90 \). This means the combined cost of materials \( M \) and labor \( L \) must equal 90.

Step 2: Solve the Constraint for a Variable

To use the constraint in the sales function, solve for one variable in terms of the other. Let's express \( M \) in terms of \( L \):\[ M = 90 - L \]

Step 3: Substitute into the Sales Equation

Substitute \( M = 90 - L \) into the sales function to eliminate \( M \):\[ S(L) = (90 - L)L - L^2 = 90L - L^2 - L^2 = 90L - 2L^2 \]

Step 4: Simplify the Substitute Equation

After substitution and simplification, we have the function\[ S(L) = 90L - 2L^2 \]which is a quadratic equation in terms of \( L \).

Step 5: Find the Vertex of the Quadratic Function

Since \( S(L) = 90L - 2L^2 \) is a quadratic equation of the form \( ax^2 + bx + c \), we identify \( a = -2 \) and \( b = 90 \). The maximum or minimum value of a quadratic function occurs at the vertex, given by \[ L = -\frac{b}{2a} = -\frac{90}{2(-2)} = 22.5 \]

Step 6: Calculate the Corresponding Maximum Value of Sales

Substitute \( L = 22.5 \) back into the expression for \( M \):\[ M = 90 - 22.5 = 67.5 \]Now, substitute \( L = 22.5 \) into the sales equation:\[ S(22.5) = 90(22.5) - 2(22.5)^2 = 90 \times 22.5 - 2 \times 506.25 = 2025 - 1012.5 = 1012.5 \]

Step 7: Conclusion of the Optimization

The maximum sales value given the budget constraint is when \( L = 22.5 \) and \( M = 67.5 \), resulting in maximum sales of \( S = 1012.5 \).

Key Concepts

Quadratic Function

A quadratic function is a mathematical expression of the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants with \(a eq 0\). These functions are characterized by their U-shaped graphs called parabolas. The orientation of the parabola depends on the sign of \(a\):

• If \(a > 0\), the parabola opens upwards.
• If \(a < 0\), the parabola opens downwards.

In our problem, the quadratic function is derived from the original sales equation \(S(L) = 90L - 2L^2\). Here, \(a\) is \(-2\), making the parabola open downwards. This is crucial because when optimizing, especially for maximum values, the downward-opening parabola informs us that the vertex gives us the maximum point of the function.

Understanding these properties and how to manipulate a quadratic function sets the baseline for further solving optimization problems in various real-world scenarios.

Vertex Formula

The vertex of a quadratic function is the point where it reaches either its maximum or minimum value, depending on the direction of the parabola. It can be calculated using the vertex formula:

\[L = -\frac{b}{2a}\]In the context of our exercise, the sales function after substitution was \(S(L) = 90L - 2L^2\), which fits the general form \(ax^2 + bx + c\) with \(a = -2\) and \(b = 90\). Applying the vertex formula:

\[L = -\frac{90}{2(-2)} = 22.5\]This tells us that the maximum sales occur when \(L\), the cost of labor, is 22.5. Why is this useful? Knowing the vertex allows us to efficiently find optimal points in problems like maximizing profits or minimizing costs. The vertex formula is thus a powerful tool in optimization, providing a straightforward method to find extremes in quadratic relationships.

Budget Constraint

A budget constraint represents the limits on the spending for various resources due to a fixed total budget. It's an essential concept in optimization problems, especially in economics and business, where resources are often limited. In the given exercise, the budget constraint is expressed as:
\[M + L = 90\]This equation limits the combined spending on materials \(M\) and labor \(L\) to 90 units. The constraint helps to determine feasible solutions that meet the financial limitations of a scenario. By solving it for one variable, for example, \(M = 90 - L\), we can isolate one variable to substitute into another relevant function.

Incorporating the budget constraint allows us to rewrite the complex sales function into a simpler quadratic form that is easier to optimize. Thus, understanding and applying budget constraints is crucial in real-world applications where multiple factors and their trade-offs need to be considered within financial boundaries.