To find the local extreme values of the function $f(x, y) = 6x + y + 4$ subject to the constraint $g(x, y) = xy = 1$ using the method of Lagrange multipliers, we follow these steps:

Step 1: Set up the Lagrange multiplier equations

We introduce a Lagrange multiplier $\lambda$ and set up the system of equations given by the gradients of the function and the constraint: $\nabla f(x, y) = \lambda \nabla g(x, y)$ This gives the following system:

• The gradient of $f(x, y)$ is $\nabla f(x, y) = (6, 1)$
• The gradient of $g(x, y) = xy - 1$ is $\nabla g(x, y) = (y, x)$

Thus, the Lagrange multiplier equations are: $(6, 1) = \lambda (y, x)$ This produces the two equations:

1. $6 = \lambda y$
2. $1 = \lambda x$

Step 2: Solve for $\lambda$

From the first equation, solve for $\lambda$: $\lambda = \frac{6}{y}$ From the second equation: $\lambda = \frac{1}{x}$ Equating the two expressions for $\lambda$, we get: $\frac{6}{y} = \frac{1}{x}$ Cross-multiply: $6x = y$

Step 3: Use the constraint

We now use the constraint $xy = 1$. Substitute $y = 6x$ into the constraint: $x(6x) = 1$ $6x^2 = 1$ $x^2 = \frac{1}{6}$ $x = \pm \frac{1}{\sqrt{6}}$

Step 4: Find corresponding values of $y$

For $x = \frac{1}{\sqrt{6}}$, we have: $y = 6x = 6 \times \frac{1}{\sqrt{6}} = \sqrt{6}$ For $x = -\frac{1}{\sqrt{6}}$, we have: $y = 6x = 6 \times -\frac{1}{\sqrt{6}} = -\sqrt{6}$

Step 5: Evaluate $f(x, y)$ at the critical points

Now, evaluate the function $f(x, y) = 6x + y + 4$ at the critical points:

1. At $\left( \frac{1}{\sqrt{6}}, \sqrt{6} \right)$: $f\left( \frac{1}{\sqrt{6}}, \sqrt{6} \right) = 6 \times \frac{1}{\sqrt{6}} + \sqrt{6} + 4 = \sqrt{6} + \sqrt{6} + 4 = 2\sqrt{6} + 4$
2. At $\left( -\frac{1}{\sqrt{6}}, -\sqrt{6} \right)$: $f\left( -\frac{1}{\sqrt{6}}, -\sqrt{6} \right) = 6 \times \left( -\frac{1}{\sqrt{6}} \right) + \left( -\sqrt{6} \right) + 4 = -\sqrt{6} - \sqrt{6} + 4 = -2\sqrt{6} + 4$

Step 6: Determine the nature of the extrema

• At $\left( \frac{1}{\sqrt{6}}, \sqrt{6} \right)$, $f(x, y) = 2\sqrt{6} + 4$
• At $\left( -\frac{1}{\sqrt{6}}, -\sqrt{6} \right)$, $f(x, y) = -2\sqrt{6} + 4$

Since the function $f(x, y) = 6x + y + 4$ is linear, it does not have global extrema on the entire plane, but the constraint $xy = 1$ restricts it to a hyperbola. Therefore, the values $2\sqrt{6} + 4$ and $-2\sqrt{6} + 4$ are local extrema, but they are not necessarily global.

Conclusion:

• The maximum value is $2\sqrt{6} + 4$
• The minimum value is $-2\sqrt{6} + 4$

Do you want further details or have any questions?

Here are 5 related questions to expand on this topic:

1. What are the conditions for global extrema in Lagrange multiplier problems?
2. How does the method of Lagrange multipliers apply to higher dimensions?
3. Can Lagrange multipliers be used with inequality constraints?
4. What role does the constraint play in determining the extrema?
5. How does linearity in $f(x, y)$ influence the outcome of Lagrange multiplier problems?

Tip: Always check the nature of the constraint when solving Lagrange multiplier problems, as it can restrict the domain of the function and affect whether the extrema are local or global.