Question

Check assumptions Consider the function \(f(x, y)=x y+x+y+100\) subject to the constraint \(x y=4\) a. Use the method of Lagrange multipliers to write a system of three equations with three variables \(x, y,\) and \(\lambda\) b. Solve the system in part (a) to verify that \((x, y)=(-2,-2)\) and \((x, y)=(2,2)\) are solutions. c. Let the curve \(C_{1}\) be the branch of the constraint curve corresponding to \(x>0 .\) Calculate \(f(2,2)\) and determine whether this value is an absolute maximum or minimum value of \(f\) over \(C_{1} \cdot(\text {Hint}: \text { Let } h_{1}(x), \text { for } x>0, \text { equal the values of } f\) over the \right. curve \(C_{1}\) and determine whether \(h_{1}\) attains an absolute maximum or minimum value at \(x=2 .\) ) d. Let the curve \(C_{2}\) be the branch of the constraint curve corresponding to \(x<0 .\) Calculate \(f(-2,-2)\) and determine whether this value is an absolute maximum or minimum value of \(f\) over \(C_{2} .\) (Hint: Let \(h_{2}(x),\) for \(x<0,\) equal the values of \(f\) over the curve \(C_{2}\) and determine whether \(h_{2}\) attains an absolute maximum or minimum value at \(x=-2 .\) ) e. Show that the method of Lagrange multipliers fails to find the absolute maximum and minimum values of \(f\) over the constraint curve \(x y=4 .\) Reconcile your explanation with the method of Lagrange multipliers.

Short answer

Answer: The absolute minimum value of the function is 108 at the point (2, 2) over curve \(C_1\) (where \(x>0\)), and the absolute maximum value is 96 at the point (-2, -2) over curve \(C_2\) (where \(x<0\)).

Step-by-step solution

Part a: Formulating the system of equations using Lagrange multipliers

First, we need to set up the Lagrangian function, \(L(x, y, \lambda)=f(x, y)-\lambda (xy-4)\) where \(f(x,y) = xy+x+y+100\). Now, to find the critical points, we'll take partial derivatives with respect to \(x, y,\) and \(\lambda\) and set them equal to zero: 1. \(\frac{\partial L}{\partial x} = y+1-\lambda y = 0\) 2. \(\frac{\partial L}{\partial y} = x+1-\lambda x = 0\) 3. \(\frac{\partial L}{\partial \lambda} = 4-xy = 0\) Now we have the system of three equations with three variables \(x, y,\) and \(\lambda\).

Part b: Solving the system of equations

We are given that the solutions are \((x, y) = (-2, -2)\) and \((x, y) = (2, 2)\). To verify these solutions, we'll substitute the values into the equations from part (a): 1. For \((-2, -2)\): - \((-1) - (2\lambda) = 0\), \(\lambda = -\frac{1}{2}\) - \((-1) - (-2\lambda) = 0\), \(\lambda = -\frac{1}{2}\) - \((-2)(-2) = 4\) All the equations are satisfied, so \((-2, -2)\) is a solution. 2. For \((2, 2)\): - \((1) - (2\lambda) = 0\), \(\lambda = \frac{1}{2}\) - \((1) - (2\lambda) = 0\), \(\lambda = \frac{1}{2}\) - \((2)(2) = 4\) All the equations are satisfied, so \((2, 2)\) is a solution.

Part c: Analyzing the curve C1

The curve \(C_1\) corresponds to the constraint curve with \(x>0\). We'll find \(f(2,2)\) and determine whether it's an absolute maximum or minimum value of \(f\) over \(C_1\). Since \(f(x,y) = xy + x + y + 100\), \(f(2, 2) = (2)(2) + 2 + 2 + 100 = 108\) To check if \(f(2, 2)\) is an absolute maximum or minimum over \(C_1\), we'll have to analyze the function \(h_1(x) = (x,\frac{4}{x} + x + \frac{4}{x} + 100\). Since \(C_1: xy = 4\) and \(x>0\), we can replace \(y\) with \(\frac{4}{x}\) in the function \(f(x, y)\). Now we'll check the first and second derivatives of \(h_1(x)\): \(h_1'(x) = 1 - \frac{4}{x^2}\) \(h_1''(x) = \frac{8}{x^3}\) Since \(h_1''(x)\) is always positive for \(x>0\), the function \(h_1(x)\) is concave up, meaning that \((2, 2)\) is an absolute minimum value of \(f\) over \(C_1\).

Part d: Analyzing the curve C2

The curve \(C_2\) corresponds to the constraint curve with \(x<0\). We'll find \(f(-2,-2)\) and determine whether it's an absolute maximum or minimum value of \(f\) over \(C_2\). Since \(f(x,y) = xy + x + y + 100\), \(f(-2, -2) = (-2)(-2) - 2 - 2 + 100 = 96\) To check if \(f(-2, -2)\) is an absolute maximum or minimum over \(C_2\), we'll have to analyze the function \(h_2(x) = (x,-\frac{4}{x} + x - \frac{4}{x} + 100\). Since \(C_2: xy = 4\) and \(x<0\), we can replace \(y\) with \(-\frac{4}{x}\) in the function \(f(x, y)\). Now we'll check the first and second derivatives of \(h_2(x)\): \(h_2'(x) = 1 + \frac{4}{x^2}\) \(h_2''(x) = -\frac{8}{x^3}\) Since \(h_2''(x)\) is always negative for \(x<0\), the function \(h_2(x)\) is concave down, meaning that \((-2, -2)\) is an absolute maximum value of \(f\) over \(C_2\).

Part e: Failure of the method of Lagrange multipliers

The method of Lagrange multipliers focuses on finding critical points, which are values where the gradient of the function \(f\) is parallel to the gradient of the constraint function. However, in this problem, the constraint curve is a disconnected curve with branches (\(C_1\) and \(C_2\)). In some cases, depending on the function \(f\) and constraint curves, the Lagrange multipliers method may fail to locate the correct critical points or to determine if the obtained solutions are absolute maxima or minima. In this problem, although we found the critical points \((2, 2)\) and \((-2, -2)\), we needed to analyze each branch separately to determine the definite nature of these points (absolute minimum or maximum) along \(C_1\) and \(C_2\).

Key concepts

Critical Points in Optimization

Understanding critical points is fundamental in optimization problems. In calculus, a critical point of a function of one variable is a point where the function's derivative is zero or undefined. When dealing with functions of two variables, such as in our example with the function f(x, y) = xy + x + y + 100, critical points occur where the partial derivatives with respect to all variables simultaneously equal zero.

These critical points can represent local maxima, local minima, or saddle points. The method of Lagrange multipliers is a powerful technique to find critical points when optimizing a function under a constraint. By introducing an auxiliary variable called the Lagrange multiplier, denoted as λ, we create a system of equations that allows us to solve for potential critical points where the constraint is active.

The exercise illustrates how to use Lagrange multipliers: first, by formulating a system of equations and then, by finding solutions that satisfy this system. It's imperative to affirm that just discovering potential critical points does not immediately classify them as maxima or minima – this requires further analysis.

Constraint Optimization

Working with constraints in optimization problems introduces a new layer of complexity. When you're seeking to optimize a function, but only within a certain set of allowable solutions defined by another equation (the constraint), you need to consider only those points that lie on the 'constraint curve' or 'constraint surface.'

In our illustrated example, the constraint is xy = 4. The curve defined by this equation limits the search for optimal values of the function f(x, y). The Lagrange multipliers method comes into play by allowing us to locate critical points without directly solving the constraint equation but rather by incorporating it into the optimization problem. The Lagrangian function, in this case, is L(x, y, λ) = f(x, y) - λ(xy - 4), where the subtraction of λ times the constraint ensures that potential solutions satisfy the constraint.

Key to constraint optimization is that sometimes you may find multiple branches or regions in your constraint (such as C₁ and C₂ in this problem), and you might need to evaluate each separately to discern the nature of the critical points on each.

Absolute Maximum and Minimum Values

In optimization, we're often interested in locating the absolute maximum and minimum values of a function, which are the highest and lowest values that the function can take on within a given domain. The difference between the absolute maximum or minimum and local extrema is crucial; the former are the extrema over the entire domain, while the latter could just be extremal within a small neighborhood.

In the exercise, after finding the critical points using Lagrange multipliers, additional steps are required to determine if these points yield absolute maximum or minimum values. This determination often involves further analysis, such as comparing the values of the function at critical points to the function's values at boundary points or endpoints of the domain.

Functions may also have to be evaluated separately along different branches of constraint curves. For example, f(x, y) needs to be analyzed on both C₁ (where x > 0) and C₂ (where x < 0) to discover if the critical points (2, 2) and (-2, -2) are absolute extrema. This requires invoking single-variable analysis on the functions h₁(x) and h₂(x), where the function f(x, y) is parameterized by x, given the constraint.