Problem 7 - Calculus and its applications

Question

Find the extremum of \(f(x, y)\) subject to the given constraint, and state whether it is a maximum or a minimum. $$ f(x, y)=2 y^{2}-6 x^{2} ; 2 x+y=4 $$

Step-by-Step Solution

Answer

The extremum is at \((x, y) = (8, -12)\) and is a minimum.

Step 1: Understand the Constraint

The constraint provided is a linear equation: \(2x + y = 4\). This constraint must be satisfied while finding the extremum of the function \(f(x, y)\).

Step 2: Define the Lagrangian Function

To use the method of Lagrange multipliers, define the Lagrangian function: \[ \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda (g(x, y) - 4) \] where \(g(x, y) = 2x+y\) (the constraint function). So, \[ \mathcal{L}(x, y, \lambda) = 2y^2 - 6x^2 + \lambda (2x + y - 4) \]

Step 3: Compute Partial Derivatives

Find the partial derivatives of \(\mathcal{L}\). \( \frac{\partial \mathcal{L}}{\partial x} = -12x + 2\lambda \)\( \frac{\partial \mathcal{L}}{\partial y} = 4y + \lambda \)\( \frac{\partial \mathcal{L}}{\partial \lambda} = 2x + y - 4 \)

Step 4: Set Partial Derivatives to Zero

Set each of the partial derivatives from Step 3 to zero:\[ -12x + 2\lambda = 0 \]\[ 4y + \lambda = 0 \]\[ 2x + y - 4 = 0 \]

Step 5: Solve the System of Equations

From the first equation, \( -12x + 2\lambda = 0 \), solve for \( \lambda \): \( \lambda = 6x \).From the second equation, \( 4y + \lambda = 0 \), substitute for \( \lambda \): \( 4y + 6x = 0 \).Substituting \( y = -\frac{3}{2}x \) into the constraint \( 2x + y = 4 \) gives \( 2x + (-\frac{3}{2}x) = 4\), simplifying to \( \frac{1}{2}x = 4 \), or \( x = 8 \).Now, \( y = -\frac{3}{2} \times 8 = -12 \).

Step 6: Determine Maximum or Minimum

The values obtained are \( x = 8 \) and \( y = -12 \). Substitute these back into the function: \( f(x, y) = 2(-12)^2 - 6(8)^2 = 288 - 384 = -96 \). Since the function decreases as one moves away from (0,0) and \(f(x, y)\) is negative, we have a minimum.

Key Concepts

Understanding Extremum in Functions

In mathematics, finding the extremum of a function involves identifying the points at which the function's value is either a maximum or a minimum. When dealing with a multivariable function like in this exercise, the goal is to determine the extremum subject to certain conditions or constraints.

This exercise involves the function \( f(x, y) = 2y^2 - 6x^2 \) with a linear constraint \( 2x + y = 4 \). The challenge is to find where this function reaches its highest or lowest value, given that it must satisfy the constraint at the same time.

In this context:

• A **maximum** value is the highest point in the region fulfilling all constraints.
• A **minimum** value is the lowest point under the same conditions.

The identified extremum, with the solution \( x = 8 \) and \( y = -12 \), creates a minimum in this example since substituting these values results in the lowest function value in the defined context.

The Role of Partial Derivatives

Partial derivatives are a fundamental tool in multivariable calculus used to describe how a function changes as one of its variables changes, while the others are held constant. For a function \( f(x, y) \), we often write \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) to denote the rates of change with respect to \( x \) and \( y \), respectively.

In Lagrange multiplier problems, partial derivatives help identify where what we call the Lagrangian's gradient is zero. This is crucial because it indicates potential points for extremum.

With the Lagrangian function \( \mathcal{L}(x, y, \lambda) = 2y^2 - 6x^2 + \lambda(2x + y - 4) \), the partial derivatives are computed to locate these stationary points:

• \( \frac{\partial \mathcal{L}}{\partial x} = -12x + 2\lambda \)
• \( \frac{\partial \mathcal{L}}{\partial y} = 4y + \lambda \)
• \( \frac{\partial \mathcal{L}}{\partial \lambda} = 2x + y - 4 \)

By setting these derivatives to zero, we impose the condition necessary to find an extremum point.

Solving Systems of Equations

In the Lagrange multipliers technique, solving a system of equations is essential to find the values of the variables that satisfy both the original function and the constraint. Once partial derivatives are set to zero, you obtain a set of equations that need to be solved simultaneously to find \( x \), \( y \), and \( \lambda \).

In our scenario, the derived system of equations is:

• \(-12x + 2\lambda = 0 \), leading to \( \lambda = 6x \)
• \(4y + \lambda = 0 \), substituting \(\lambda\) gives \(4y + 6x = 0 \)
• The constraint \(2x + y = 4\)

Solving this system involves substituting one equation into another to find the variable values.

Here, after appropriate substitutions, the solution found is \( x = 8 \) and \( y = -12 \). This solution meets both the function's balance and the constraint applied, thus providing the conditions for the extremum of the function.