Questions: This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x, y) = 6x + 2y; x^2 + y^2 = 10 maximum value minimum value

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint.
f(x, y) = 6x + 2y; x^2 + y^2 = 10
maximum value
minimum value

Transcript text: This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. \[ f(x, y)=6 x+2 y ; \quad x^{2}+y^{2}=10 \] maximum value $\square$ minimum value $\square$

Solution

Solution Steps

To solve this problem using Lagrange multipliers, we need to find the points where the gradients of the function $ f(x, y) $ and the constraint $ g(x, y) = x^2 + y^2 - 10 $ are parallel. This involves setting up the system of equations given by the gradients and solving for $ x $ and $ y $.

Step 1: Define the Function and Constraint

We are given the function $ f(x, y) = 6x + 2y $ and the constraint $ g(x, y) = x^2 + y^2 - 10 = 0 $.

Step 2: Compute the Gradients

The gradients of the function and the constraint are: \[ \nabla f = \begin{pmatrix} 6 \\ 2 \end{pmatrix}, \quad \nabla g = \begin{pmatrix} 2x \\ 2y \end{pmatrix} \]

Step 3: Set Up the System of Equations

Using Lagrange multipliers, we set up the system of equations: \[ \begin{cases} 6 - 2x\lambda = 0 \\ 2 - 2y\lambda = 0 \\ x^2 + y^2 - 10 = 0 \end{cases} \]

Step 4: Solve the System of Equations

Solving the system, we find the solutions: \[ (x, y, \lambda) = (-3, -1, -1) \quad \text{and} \quad (3, 1, 1) \]

Step 5: Evaluate the Function at the Solutions

Evaluating $ f(x, y) $ at the solutions: \[ f(-3, -1) = 6(-3) + 2(-1) = -18 - 2 = -20 \] \[ f(3, 1) = 6(3) + 2(1) = 18 + 2 = 20 \]