To solve this problem, we need to find the minimum and maximum values of the function:

$f(x, y) = e^{-\frac{xy}{16}}$

subject to the constraint:

$11x^2 + 14y^2 = 13$

Approach

We will use the method of Lagrange multipliers to find the extrema of $f(x, y)$ subject to the constraint.

1. Define the Lagrange function: $\mathcal{L}(x, y, \lambda) = e^{-\frac{xy}{16}} + \lambda (11x^2 + 14y^2 - 13)$

2. Take partial derivatives with respect to $x$, $y$, and $\lambda$, and set them equal to zero: $\frac{\partial \mathcal{L}}{\partial x} = 0, \quad \frac{\partial \mathcal{L}}{\partial y} = 0, \quad \frac{\partial \mathcal{L}}{\partial \lambda} = 0$

3. Solve the resulting system of equations to find the values of $x$, $y$, and $\lambda$ that satisfy the constraint.

4. Substitute these values into $f(x, y)$ to determine the minimum and maximum values.

Calculations

Let's work through these steps systematically.

Step 1: Partial Derivatives

Since this problem involves solving a system of equations, I will calculate these derivatives and solve the system.The solutions for $(x, y, \lambda)$ that satisfy the constraint are:

1. $x = -\frac{\sqrt{286}}{22}, \; y = -\frac{\sqrt{91}}{14}$
2. $x = \frac{\sqrt{286}}{22}, \; y = \frac{\sqrt{91}}{14}$
3. $x = -\frac{\sqrt{286}}{22}, \; y = \frac{\sqrt{91}}{14}$
4. $x = \frac{\sqrt{286}}{22}, \; y = -\frac{\sqrt{91}}{14}$

Next, we substitute each pair $(x, y)$ into $f(x, y) = e^{-\frac{xy}{16}}$ to determine the values of $f(x, y)$ for each case, which will give us the minimum and maximum values.The evaluated values of $f(x, y)$ at the given points are:

• $f(x, y) \approx 0.9678$ for the points $\left(-\frac{\sqrt{286}}{22}, -\frac{\sqrt{91}}{14}\right)$ and $\left(\frac{\sqrt{286}}{22}, \frac{\sqrt{91}}{14}\right)$.
• $f(x, y) \approx 1.0333$ for the points $\left(-\frac{\sqrt{286}}{22}, \frac{\sqrt{91}}{14}\right)$ and $\left(\frac{\sqrt{286}}{22}, -\frac{\sqrt{91}}{14}\right)$.

Results

• Minimum value of $f(x, y)$ is approximately $f_{\text{min}} = 0.9678$.
• Maximum value of $f(x, y)$ is approximately $f_{\text{max}} = 1.0333$.

Would you like further details on the Lagrange multiplier method, or have any additional questions?

Here are some related questions to consider:

1. What are the general conditions for using the Lagrange multiplier method?
2. How does the exponential function affect the extrema in constrained optimization?
3. What happens to the function values if the constraint changes?
4. Can we apply the Lagrange multiplier method to three-variable functions?
5. How would we approach this if the constraint equation was not quadratic?

Tip: When solving constrained optimization problems, always verify that the solutions satisfy both the original constraint and the extremum conditions.