Problem 7 - College Algebra and Calculus: An Applied Approach

Question

Use Lagrange multipliers to find the given extremum. In each case, assume that \(x\) and \(y\) are positive. $$ \text { Maximize } f(x, y)=2 x+2 x y+y \quad 2 x+y=100 $$

Step-by-Step Solution

Answer

The solution to this problem requires a mathematical approach involving forming the Lagrange function, differentiating it with respect to its variables, solving the resulting system of equations, and substituting the resultant values into the original function to get the maximum value. If successfully carried out, these steps will provide the necessary solution.

Step 1: Formulating the Lagrange Function

Create a function \(L(x, y, \lambda)\), called the Lagrange function, which incorporates the given function and the constraint. This is accomplished by using the equation: \(L(x,y,\lambda) = f(x,y) - \lambda(g(x,y) -c)\), where \(f(x,y)\) is the target function, \(g(x,y)\) is the constraint function, and \(c\) is the value of the constraint. In this exercise, the Lagrange function becomes: \(L(x, y, \lambda) = 2x + 2xy + y - \lambda (2x + y -100)\)

Step 2: Differentiating the Lagrange Function

Differentiate the Lagrange function with respect to \(x\), \(y\), and \(\lambda\) and set the derivatives equal to zero to derive a system of equations. This gives: \(\frac{\partial L}{\partial x}=2 + 2y - 2\lambda = 0\), \(\frac{\partial L}{\partial y}=2x + 1 - \lambda = 0\), and \(\frac{\partial L}{\partial \lambda}=2x + y - 100 = 0\).

Step 3: Solving the System of Equations

Solve the system of equations obtained in the previous step. The solution to the system provides the values for \(x\), \(y\), and \(\lambda\) which will be used to calculate the maximum value of the function.

Step 4: Identify Maximum

Substitute the solutions for \(x\) and \(y\) into the original function to get the maximum.

Key Concepts

Extremum Optimization

Optimizing a function to find its maximum or minimum value is a fundamental problem in mathematics and economics. It involves finding the highest or lowest point on a graph, known as the extremum. When the function is subject to constraints, the challenge intensifies. It's like trying to reach the highest or lowest step on a staircase without stepping beyond the edges.

In the given exercise, the objective is to maximize the function \( f(x, y)=2x+2xy+y \) subject to the constraint \( 2x+y=100 \). To visualize, imagine you have a landscape with hills and valleys, and you want to find the highest peak, but you must stay on a path defined by the constraint. This is where Lagrange multipliers come into play, allowing us to explore the peaks along the constrained path and find the optimal point. In this example, identifying the maximum value is akin to finding the best combination of \(x\) and \(y\) that gives you the highest point on the \(f\) graph while staying within the limits imposed by the equation \(2x+y=100\).

Lagrange Function

The Lagrange function is a brilliant mathematical invention that elegantly unifies the objective function and the constraint into a single equation. It's like having a Swiss Army knife for optimization problems. With this tool, difficult constraints become manageable.

In step 1 of the given solution, the Lagrange function is formulated as \(L(x, y, \lambda) = 2x + 2xy + y - \lambda (2x + y -100)\). Here, \(\lambda\) is the ingenious Lagrange multiplier, a new variable introduced to balance the scales between the objective function and the constraint. This auxiliary variable serves as a weight adjusting the impact of the constraint on the overall equation.

As an instructor, I would liken \(\lambda\) to a dial that can tune a guitar string. Just as you adjust the dial to reach the perfect pitch, you alter \(\lambda\) to harmonize the objective function with the constraint. The beauty of the Lagrange function lies in its ability to convert a constrained optimization problem into a solvable system of equations, paving the way toward finding those elusive optimal values of \(x\) and \(y\).

System of Equations

Following the establishment of the Lagrange function, the journey to the extremum leads us to a system of equations. This step involves setting up an intricate 'dance' of derivatives, choreographed to tune in to the subtle changes in our function as constrained by the beloved \(\lambda\).

Differentiating \(L(x, y, \lambda)\) with respect to \(x\), \(y\), and \(\lambda\) and setting these derivatives equal to zero spawns this system of equations: \[\frac{\partial L}{\partial x}=2 + 2y - 2\lambda = 0,\frac{\partial L}{\partial y}=2x + 1 - \lambda = 0,\frac{\partial L}{\partial \lambda}=2x + y - 100 = 0\].

These equations are like clues in a treasure hunt, each one leading to pieces of information that, when solved together, reveal the location of the 'treasure'—the extremum. By carefully analyzing and solving this system, just as you would piece together fragments of a map, you reveal the coordinates (values for \(x\), \(y\), and \(\lambda\)) that pinpoint the optimal point on the landscape of our function. It is an exercise in precision and problem-solving, requiring equal parts mathematical skill and detective-like logic.