Problem 8 - Calculus an Applied Approach

Question

Use Lagrange multipliers to find the given extremum. In each case, assume that \(x\) and \(y\) are positive. $$ \begin{array}{ll}{\text { Objective Function }} & {\text { Constraint }} \\\ {\text { Maximize } f(x, y)=3 x+y+10} & {x^{2} y=6}\end{array} $$

Step-by-Step Solution

Answer

Compute first the Lagrangian and find its derivative with respect to each variable. Use the equations obtained to find the extremum. Then, substitute these extremum points back into the objective function to find the maximum value of the function.

Step 1: Setting up the Lagrangian function

The first step is to set up the Lagrangian function. The Lagrangian function is defined as \(L(x, y, \lambda) = f(x, y) - \lambda (g(x, y) - c)\), where \(f(x, y)\) is the objective function, \(g(x, y)\) is the constraint function, and \(\lambda\) is the Lagrange multiplier. Here, the Lagrangian function is \(L(x, y, \lambda) = 3x + y + 10 - \lambda (x^{2}y - 6)\).

Step 2: Take derivative and set equal to zero

The next step is to take the derivative of the Lagrangian function with respect to each variable and set each resulting equation equal to zero to find the critical points. It gives us three equations: \\(1) \\frac{\partial L}{\partial x} = 3 - 2\lambda xy - \lambda y^{2} = 0, \\(2) \\frac{\partial L}{\partial y} = 1 - \lambda x^{2} = 0, and \\(3) \\frac{\partial L}{\partial \lambda} = x^{2}y - 6 = 0.

Step 3: Solve the equations

The three resulting equations can now be solved. From equation (2), we can solve for \(\lambda\): \(\lambda = \frac{1}{x^{2}}\). We can then substitute \(\lambda\) into equation (1) to get \(3 - 2\frac{y}{x} - \frac{y^{2}}{x^{2}} = 0\). Solving this gives two possible values for \(y\). Then, substitute \(y\) into equation (3) to get the corresponding \(x\) values.

Step 4: Find the maximum value

To find the maximum value, substitute the \(x\) and \(y\) values obtained in the previous step back into the objective function \(f(x, y) = 3x + y + 10\). The maximum value will be the highest out of all the values calculated.

Key Concepts

Objective Function

When solving an optimization problem using Lagrange multipliers, the objective function is the key target to be maximized or minimized. In the given exercise, the objective function is \( f(x, y) = 3x + y + 10 \).

This function tells us the quantity we want to optimize, depending on variables \(x\) and \(y\). Here, we aim to find the maximum value of this function, which is influenced by the inputs \(x\) and \(y\). It's crucial to understand that without any constraints, this function would increase indefinitely with larger values of \(x\) and \(y\).

However, practical problems usually have constraints that restrict feasible solutions. Hence, the Lagrange multipliers method helps in maximizing or minimizing this function under specified conditions.

Constraint Function

Constraints are the rules or conditions that our solution must satisfy. In this exercise, the constraint function is given as \( x^2 y = 6 \).

The constraint connects variables \(x\) and \(y\) in a specific relationship. Unlike the objective function, constraints do not seek to be optimized. Instead, they define the boundary within which the solution should reside.

• Constraints ensure the solutions are practical and adhere to real-world limitations.
• They transform an unlimited problem into a feasible one, making optimization meaningful in practical scenarios.

In our example, we can imagine this constraint as a surface or a curve within the \((x, y)\) space, limiting the points \(x\) and \(y\) to lie on this surface.

Critical Points

Critical points are essential for determining where our objective function satisfies the constraint function in the best way possible. To find critical points using Lagrange multipliers, we first express these as intersections of gradients.

This involves calculating derivatives and setting them to zero. In our case, after setting up the Lagrangian, derivatives with respect to \(x\), \(y\), and \(\lambda\) are computed.
Solving these equations can identify potential maximum or minimum values that respect our constraint.

• Critical points are solutions to equations derived from the Lagrangian function.
• They represent candidates for optimal points of the objective function.

Hence, identifying all critical points is crucial to ensure the accurate solution of optimization problems.

Lagrangian Function

The Lagrangian function is a core component in solving constrained optimization problems with Lagrange multipliers. It incorporates both the objective function and the constraint function.

Specifically, the Lagrangian combines them into one equation: \[ L(x, y, \lambda) = f(x, y) - \lambda (g(x, y) - c) \] Here,

• \( f(x, y) \) is the objective function.
• \( g(x, y) - c \) represents the constraint.
  • This part is multiplied by \(\lambda\), the Lagrange multiplier.

The multiplier \(\lambda\) reveals how much the objective function would improve if the constraint were eased.

During the optimization process, forming the Lagrangian is crucial as it facilitates finding critical points and thereby the optimized solution.