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Use Lagrange multipliers to find the maximum and minimum values of \(f\) (when they exist) subject to the given constraint. $$f(x, y)=x y \text { subject to } x^{2}+y^{2}-x y=9$$

Short Answer

Expert verified
Answer: The maximum and minimum values of the function f(x, y) = xy under the given constraint are both 0.

Step by step solution

01

Find the gradient of f(x, y) and g(x, y)

Find the gradient of the function f(x, y) = xy and the constraint function g(x, y) = x^2 + y^2 - xy - 9. The gradient is the partial derivative of the function with respect to x and y. For f(x, y): \(\frac{\partial f}{\partial x} = y\) \(\frac{\partial f}{\partial y} = x\) For g(x, y): \(\frac{\partial g}{\partial x} = 2x - y\) \(\frac{\partial g}{\partial y} = 2y - x\)
02

Set up the Lagrange equation

Now, we will set up the Lagrange equation. We introduce a new variable called the Lagrange multiplier (λ). The equations are set up as follows: \(\frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x}\) \(\frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}\) \(g(x, y) = 0\) Plugging in the values we obtained for the gradients in step 1, we get: \(y = \lambda (2x - y)\) \(x = \lambda (2y - x)\) \(x^2 + y^2 - xy = 9\)
03

Solve for the stationary points

Now, we need to solve the system of equations obtained in Step 2 to find values for x, y, and λ. Rearranging the first and second equations to make x and y the subjects: \(x= \frac{\lambda}{\lambda+1}(2y-x)\) \(y= \frac{\lambda}{\lambda+1}(2x-y)\) Now substitute the expression for x into the constraint equation: \(\left(\frac{\lambda}{\lambda+1}(2y-x)\right)^2 + y^2 - \left(\frac{\lambda}{\lambda+1}(2y-x)\right)y = 9\) Similarly, substitute the expression for y into the constraint equation: \(x^2 + \left(\frac{\lambda}{\lambda+1}(2x-y)\right)^2 - x\left(\frac{\lambda}{\lambda+1}(2x-y)\right) = 9\) Solve these equations for x, y, and λ. After solving, we get 4 sets of (x, y, λ) values: \((0, -3, \frac{1}{2})\) \((0, 3, -2)\) \((-3, 0, 2)\) \((3, 0, \frac{1}{2})\)
04

Find the maximum and minimum values of f(x, y)

Finally, we need to plug in the stationary points obtained in step 3 into the original function f(x, y) = xy to find the maximum and minimum values. \(f(0, -3) = (0)(-3) = 0\) \(f(0, 3) = (0)(3) = 0\) \(f(-3, 0) = (-3)(0) = 0\) \(f(3, 0) = (3)(0) = 0\) In this case, all values are equal to 0. So, the maximum and minimum values of the function f(x, y) = xy under the given constraint are both 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gradient
In mathematics, the concept of a gradient is essential, especially when working with functions that have more than one variable. Imagine the gradient as a kind of compass that tells us the steepest direction to move at any point on the surface of the function.
The gradient is represented as a vector consisting of the partial derivatives of the function with respect to each of its variables.
  • For example, the gradient of a function \(f(x, y)\) is \(abla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\).
In the context of the exercise, the function \(f(x, y) = xy\) had its gradient calculated as \(\left(y, x\right)\). Similarly, for the constraint \(g(x, y) = x^2 + y^2 - xy - 9\), the gradient is \(\left(2x - y, 2y - x\right)\).
Understanding the gradient helps in analyzing how the function behaves and changes in different directions. It's especially useful in optimization tasks like finding the maximum or minimum values of a function.
Constraint Optimization
Constraint optimization is a process used when we have to find the best solution under a given restriction or condition. This is common in real-world problems where certain variables are limited, like finding the maximum production rate given limited resources.
In the exercise, Lagrange multipliers come into play to handle the constraint optimization. The tricky part is to optimize the function \(f(x, y) = xy\) while adhering to the constraint \(x^2 + y^2 - xy = 9\).
Lagrange multipliers introduce a new variable, \(\lambda\), to form equations that allow us to solve for the stationary points. These equations ensure our solution meets the constraint while optimizing \(f(x, y)\).
  • The steps involve setting the gradients of \(f(x, y)\) and the constraint equal, scaled by this multiplier \(\lambda\).
  • Then, solving these equations and the original constraint equation helps us find the values that satisfy all conditions.
The output lets us determine if we are dealing with a maximum, minimum, or saddle point of the function under the constraint.
Stationary Points
Stationary points are crucial in determining where a function's rate of change is zero in particular directions. This idea is key to finding maximum or minimum values in nonlinear programming or calculus.
  • In mathematical terms, the stationary points occur when the gradient of the function equals zero. For a constrained function, this means solving the system set up by the gradients and Lagrange multipliers.
In the exercise, solving the system of equations involving \(x, y,\) and \(\lambda\) yielded possible solutions that were then checked to see how they affected the function \(f(x, y) = xy\).
After plugging these points into the original function, we discovered that all provided outcomes led to the same value: 0. Identifying these stationary points and their corresponding function values helps us to determine the optimal scenarios under the given constraints.
In many cases, these points can tell us where an actual maximum or minimum might occur, or if there are simply no meaningful optimal values based on the constraints given. This shows how powerful the method of Lagrange multipliers is in the context of constraint optimization.

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