Problem 7 - The Calculus with Analytic Geometry

Question

In Exercises 7 through 12 , use the method of Lagrange multipliers to find the critical points of the given function subject to the indicated constraint. \(f(x, y)=x^{2}+2 x y+y^{2}\) with constraint \(x-y=3\)

Step-by-Step Solution

Answer

Critical point: \((\frac{3}{2}, -\frac{3}{2})\); Function value: \(\frac{9}{4}\)

Step 1: Define the function and constraint

Given function: \[ f(x, y) = x^2 + 2xy + y^2 \]Constraint: \[ g(x, y) = x - y - 3 = 0 \]

Step 2: Set up the Lagrange function

Form the Lagrangian: \[ \mathcal{L}(x, y, \lambda) = f(x, y) - \lambda g(x, y) \] This gives us: \[ \mathcal{L}(x, y, \lambda) = x^2 + 2xy + y^2 - \lambda(x - y - 3) \]

Step 3: Find partial derivatives

Compute the partial derivatives and set them equal to zero: \[ \frac{\partial \mathcal{L}}{\partial x} = 2x + 2y - \lambda = 0 \]\[ \frac{\partial \mathcal{L}}{\partial y} = 2x + 2y + \lambda = 0 \]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = -(x - y - 3) = 0 \]

Step 4: Solve the system of equations

From \[ \frac{\partial \mathcal{L}}{\partial x} = 2x + 2y - \lambda = 0 \] and \[ \frac{\partial \mathcal{L}}{\partial y} = 2x + 2y + \lambda = 0 \], add them to get:\[ 2x + 2y - \lambda + 2x + 2y + \lambda = 0 \Rightarrow 4x + 4y = 0 \Rightarrow x + y = 0 \] and from \[ x - y - 3 = 0 \Rightarrow x = y + 3 \]. Combining the two equations \[ x + y = 0 \] and \[ x = y + 3 \] gives us: \[ y + 3 + y = 0 \Rightarrow 2y + 3 = 0 \Rightarrow y = -\frac{3}{2} \] and\[ x = \frac{3}{2} \]

Step 5: Check the solution in the constraint

Put the values \[ x = \frac{3}{2} \] and \[ y = -\frac{3}{2} \] back into the constraint to verify:\[ x - y = \frac{3}{2} + \frac{3}{2} = 3 \]. Hence verified.

Step 6: Find the function value at the critical point

Calculate \[ f(x, y) = f(\frac{3}{2}, -\frac{3}{2}) = (\frac{3}{2})^2 + 2(\frac{3}{2})(-\frac{3}{2}) + (-\frac{3}{2})^2 = \frac{9}{4} - \frac{9}{4} + \frac{9}{4} = \frac{9}{4} \]

Key Concepts

Critical Points

Critical points are values of variables at which the function you are examining reaches a local maximum, minimum, or saddle point. In your problem, you are finding critical points for a function subjected to a constraint. By solving the partial derivatives and satisfying the given constraint, you identify points that could be potential maxima, minima, or saddle points.

For our function, the critical points are where the partial derivatives of the Lagrangian are zero, and the constraint is satisfied. These points provide insights into the behavior of the function under the given restrictions.

Constraint Optimization

Constraint optimization is a method where you maximize or minimize a function subject to one or more restrictions (constraints). In this exercise, your function \(f(x, y)\) is optimized under the constraint \(x-y=3\).

Using Lagrange multipliers, you convert this constrained problem into an unconstrained one, allowing you to find the extrema more straightforwardly. You set up a Lagrangian function incorporating the constraint and take its partial derivatives. This powerful technique simplifies finding optimal points when faced with limitations.

Multivariable Calculus

Multivariable calculus involves functions of more than one variable. It extends the concepts of single-variable calculus to higher dimensions. In this exercise, we've dealt with a function \(f(x, y)\) that has two variables, \(x\) and \(y\).

Multivariable calculus requires you to understand partial derivatives, gradients, and how to find critical points in higher dimensions. Setting the partial derivatives to zero is crucial for finding these points, and Lagrange multipliers help when constraints are involved.

Understanding how to operate in higher dimensions allows you to solve complex real-world problems, from physics to economics.

Lagrangian Function

The Lagrangian function combines the original function and the constraints using a multiplier, \(\lambda\). In this method, you modify your function \(f(x, y)\) to \(\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda g(x, y)\).

In this exercise:

• Original function: \(f(x, y) = x^2 + 2xy + y^2\)
• Constraint: \(g(x, y) = x - y - 3 = 0\)
• Lagrangian: \mathcal{L}(x, y, \lambda) = x^2 + 2xy + y^2 - \lambda (x - y - 3)\

By formulating the Lagrangian, you transform a constrained problem into a format where you can apply standard optimization techniques. This makes handling constraints more manageable and solution-oriented.

Partial Derivatives

Partial derivatives measure how a function changes as its variables change. They are essential in finding the critical points of functions with multiple variables. When constructing the Lagrangian, you take partial derivatives with respect to each variable and set them to zero.

In the exercise, we calculated the partial derivatives:

• \( \frac{\partial \mathcal{L}}{\partial x} = 2x + 2y - \lambda = 0 \)
• \( \frac{\partial \mathcal{L}}{\partial y} = 2x + 2y + \lambda = 0 \)
• \( \frac{\partial \mathcal{L}}{\partial \lambda} = -(x - y - 3) = 0 \)

These equations are then solved to get the values of \(x, y,\) and \(\lambda\) that satisfy both the function and the constraint. Partial derivatives give you a tool to explore the behavior of multivariable functions comprehensively.