We are tasked with finding the extreme values of $f(x, y) = x^2 + 2y^2$ subject to the constraint $g(x, y) = 2x + y = 3$.

This is a typical constrained optimization problem, and we will use the method of Lagrange multipliers to solve it.

Step 1: Define the Lagrange Multiplier System

We introduce a Lagrange multiplier, $\lambda$, and solve the system: $\nabla f = \lambda \nabla g$ where:

• $f(x, y) = x^2 + 2y^2$
• $g(x, y) = 2x + y - 3 = 0$ (rearranged the constraint to equal 0)

First, compute the gradients: $\nabla f = (2x, 4y)$ $\nabla g = (2, 1)$

Now, set up the system of equations:

1. $2x = \lambda \cdot 2$
2. $4y = \lambda \cdot 1$
3. The constraint $2x + y = 3$

Step 2: Solve the system of equations

From the first equation, solve for $\lambda$: $\lambda = x$

From the second equation: $4y = \lambda \quad \Rightarrow \quad 4y = x \quad \Rightarrow \quad x = 4y$

Substitute $x = 4y$ into the constraint equation $2x + y = 3$: $2(4y) + y = 3 \quad \Rightarrow \quad 8y + y = 3 \quad \Rightarrow \quad 9y = 3 \quad \Rightarrow \quad y = \frac{1}{3}$

Now, substitute $y = \frac{1}{3}$ into $x = 4y$: $x = 4\left(\frac{1}{3}\right) = \frac{4}{3}$

Step 3: Compute the extreme values

Now that we have $x = \frac{4}{3}$ and $y = \frac{1}{3}$, we substitute these values into the function $f(x, y) = x^2 + 2y^2$: $f\left(\frac{4}{3}, \frac{1}{3}\right) = \left(\frac{4}{3}\right)^2 + 2\left(\frac{1}{3}\right)^2 = \frac{16}{9} + 2\left(\frac{1}{9}\right) = \frac{16}{9} + \frac{2}{9} = \frac{18}{9} = 2$

Thus, the minimum value of $f(x, y)$ is 2.

Since there are no other solutions and no maximum is specified by the problem, the maximum does not exist (DNE).

Final Answer:

• Maximum: DNE
• Minimum: 2

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. What happens if the constraint equation is different, such as $g(x, y) = 2x + 3y = 5$?
2. Can this method be applied if $f(x, y)$ involves higher powers or different functions, like trigonometric terms?
3. How does the method of Lagrange multipliers change when there are multiple constraints?
4. Can the method of Lagrange multipliers be used for maximizing a function with several variables?
5. How would the problem change if we had inequality constraints rather than equality constraints?

Tip: The Lagrange multiplier $\lambda$ can be interpreted as the rate at which the function $f(x, y)$ increases with respect to changes in the constraint $g(x, y)$.