Question

Use the Lagrange-multiplier method to find the stationary values of \(z\) : \((a) z=x y,\) subject to \(x+2 y=2\) (b) \(z=x(y+4),\) subject to \(x+y=8\) (c) \(z=x-3 y-x y,\) subject to \(x+y=6\) \((d) z=7-y+x^{2},\) subject to \(x+y=0\)

Short answer

Stationary values are: (a) 0.5, (b) 36, (c) -3, (d) 7.

Step-by-step solution

Understand the Lagrange Multiplier Method

The Lagrange Multiplier method is a strategy for finding the local maxima and minima of a function subject to equality constraints. Given a function to be maximized or minimized, say \( z(x, y) \), and a constraint \( g(x, y) = 0 \), we introduce a multiplier \( \lambda \) and solve the system of equations \( abla z = \lambda abla g \) and \( g(x, y) = 0 \).

Define the functions for part (a)

For \( z = x y \) with constraint \( x + 2y = 2 \), the functions are: - Function to optimize: \( z = x y \)- Constraint: \( g(x, y) = x + 2y - 2 = 0 \)

Compute the gradients for part (a)

Determine the gradients of \( z \) and \( g \):- \( abla z = \left( y, x \right) \)- \( abla g = \left( 1, 2 \right) \)

Set up Lagrange equations for part (a)

Using \( abla z = \lambda abla g \), we obtain:- \( y = \lambda \)- \( x = 2\lambda \)- Along with constraint: \( x + 2y = 2 \)

Solve equations for part (a)

From \( y = \lambda \) and \( x = 2\lambda \), substitute into the constraint:- \( 2\lambda + 2\lambda = 2 \rightarrow 4\lambda = 2 \rightarrow \lambda = \frac{1}{2} \)- \( x = 1, y = \frac{1}{2} \)- Stationary value of \( z = x y = 1 \cdot \frac{1}{2} = \frac{1}{2} \)

Define the functions for part (b)

For \( z = x(y + 4) \) with constraint \( x + y = 8 \), the functions are:- Function to optimize: \( z = x(y + 4) \)- Constraint: \( g(x, y) = x + y - 8 = 0 \)

Compute the gradients for part (b)

Determine the gradients of \( z \) and \( g \):- \( abla z = \left( y + 4, x \right) \)- \( abla g = \left( 1, 1 \right) \)

Set up Lagrange equations for part (b)

Using \( abla z = \lambda abla g \), we obtain:- \( y + 4 = \lambda \)- \( x = \lambda \)- Along with constraint: \( x + y = 8 \)

Solve equations for part (b)

From \( x = \lambda \) and \( y + 4 = \lambda \), substitute into the constraint:- \( \lambda + \lambda - 4 = 8 \rightarrow 2\lambda = 12 \rightarrow \lambda = 6 \)- \( x = 6, y = 2 \)- Stationary value of \( z = 6(2 + 4) = 36 \)

Define the functions for part (c)

For \( z = x - 3y - xy \) with constraint \( x + y = 6 \), the functions are:- Function to optimize: \( z = x - 3y - xy \)- Constraint: \( g(x, y) = x + y - 6 = 0 \)

Compute the gradients for part (c)

Determine the gradients of \( z \) and \( g \):- \( abla z = \left( 1 - y, -3 - x \right) \)- \( abla g = \left( 1, 1 \right) \)

Set up Lagrange equations for part (c)

Using \( abla z = \lambda abla g \), we obtain:- \( 1 - y = \lambda \)- \( -3 - x = \lambda \)- Along with constraint: \( x + y = 6 \)

Solve equations for part (c)

Equating both expressions for \( \lambda \):- \( 1 - y = -3 - x \rightarrow x - y = 4 \)- Solve with constraint: \( x + y = 6 \)- Add equations: \( 2x = 10 \rightarrow x = 5 \), \( y = 1 \)- Stationary value of \( z = 5 - 3(1) - 5(1) = -3 \)

Define the functions for part (d)

For \( z = 7 - y + x^2 \) with constraint \( x + y = 0 \), the functions are:- Function to optimize: \( z = 7 - y + x^2 \)- Constraint: \( g(x, y) = x + y = 0 \)

Compute the gradients for part (d)

Determine the gradients of \( z \) and \( g \):- \( abla z = \left( 2x, -1 \right) \)- \( abla g = \left( 1, 1 \right) \)

Set up Lagrange equations for part (d)

Using \( abla z = \lambda abla g \), we obtain:- \( 2x = \lambda \)- \( -1 = \lambda \)- Along with constraint: \( x + y = 0 \)

Solve equations for part (d)

Since \( 2x = \lambda \) and \( \lambda = -1 \), we have \( 2x = -1 \rightarrow x = -\frac{1}{2} \).- Using the constraint: \( x + y = 0 \rightarrow y = \frac{1}{2} \)- Stationary value of \( z = 7 - \frac{1}{2} + \left(-\frac{1}{2}\right)^2 = 7 \)

Conclusion

The stationary values of \( z \) for each part are found by solving the Lagrange equations combined with the constraints. These points give us potential maxima, minima, or saddle points subject to the given conditions.

Key concepts

Stationary Values

When using the Lagrange Multiplier Method, we aim to find the stationary values of a function. Stationary values are points where the function has a derivative of zero, which indicates potential maxima, minima, or saddle points. In the context of constrained optimization, these points satisfy the Lagrangian equations derived from the function of interest and the given constraints.

In exercises like the ones provided, stationary values are calculated by equating the gradients of the original function and the constraint multiplied by the Lagrange Multiplier, \( \lambda \). Once we have solved these equations together with the constraint equation, we get specific values of the variables where the function could have one of these extreme points.

The stationary value, however, needs verification for whether it is a maximum, minimum, or a saddle point using the second derivative test or other methods, such as examining the behavior of the function around these points.

Equality Constraints

Equality constraints are conditions that must be satisfied for a particular problem. In an optimization problem, these constraints shape the feasible region — the set of points that are considered when searching for a function's optimal value. The constraints are typically expressed in the form \( g(x, y) = 0 \).

The presence of equality constraints implies that we cannot freely choose any values for the variables, but have to respect these specific conditions that define possible solutions. In the exercises provided, each part had a unique constraint that ensured the variables met shapes like lines or planes, narrowing down the area where optimization can be correctly achieved.

The constraint works hand in hand with the function. The feasible solution must meet both the conditions of the problem and provide the optimum value for what is being calculated, thereby satisfying all conditions simultaneously.

Gradient Calculation

Calculating gradients is a vital step in using the Lagrange Multiplier Method. A gradient is a vector that points in the direction of the steepest ascent of a function. It consists of partial derivatives concerning each variable within the function.

To compute gradients, we differentiate the function concerning each of its variables. For instance, if a function \( f(x, y) \) is given, its gradient, \( abla f \), would be \( (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \). Similarly, we compute the gradient of the constraint \( g(x, y) \).

The calculated gradient of the objective function equated with the product of a scalar (Lagrange multiplier) and the gradient of the constraint enables the determination of where the function's steepest path is blocked by the constraint. Thus, gradients help identify the points that are potential solutions to the optimization problem.

Optimization Problem

An optimization problem entails finding the maximum or minimum value of a function subject to specific constraints. In many cases, the Lagrange Multiplier Method efficiently finds these optimal points when the function and constraints are differentiable and lie in some coordinate space.

Each optimization problem provides a point or solution which represents the best value (either maximum or minimum) that the objective function can take within the given constraints. In the exercises covered, each problem required maximizing or minimizing expressions like \( z = xy \), \( z = x(y+4) \), and others, subject to linear constraints such as \( x + 2y = 2 \).

The process encompasses introducing the Lagrange multiplier, setting up equations from calculated gradients, and solving for variable values. This results in identifying potential solutions that satisfy both the optimization criteria of extreme values and the constraints imposed. These steps make it feasible to solve complex optimization problems that arise in real-world applications.