x^2 + y^2 = 1
and then f(x,y) = 2x^2 + 3y^2
do lagrange and find min max

To find the extrema of the function \( f(x, y) = 2x^2 + 3y^2 \) subject to the constraint \( g(x, y) = x^2 + y^2 - 1 = 0 \) using Lagrange multipliers, we proceed as follows:

1. **Set up the equations**:
   We want to find the points where the gradients of \( f \) and \( g \) are proportional:
   \[
   \nabla f = \lambda \nabla g
   \]

   Calculating the gradients:
   - \( \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (4x, 6y) \)
   - \( \nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) = (2x, 2y) \)

2. **Set up the system of equations**:
   From \( \nabla f = \lambda \nabla g \), we have:
   \[
   4x = \lambda (2x)
   \]
   \[
   6y = \lambda (2y)
   \]

   Additionally, we have the constraint:
   \[
   x^2 + y^2 = 1
   \]

3. **Analyze the equations**:
   From the first equation, if \( x \neq 0 \):
   \[
   \lambda = \frac{4}{2} = 2
   \]

   From the second equation, if \( y \neq 0 \):
   \[
   \lambda = \frac{6}{2} = 3
   \]

   Since these two values of \( \lambda \) are not equal, we consider the cases where either \( x = 0 \) or \( y = 0 \).

4. **Case 1: \( x = 0 \)**:
   Substituting into the constraint:
   \[
   0^2 + y^2 = 1 \implies y = \pm 1
   \]
   Then,
   \[
   f(0, 1) = 2(0)^2 + 3(1)^2 = 3
   \]
   \[
   f(0, -1) = 2(0)^2 + 3(-1)^2 = 3
   \]

5. **Case 2: \( y = 0 \)**:
   Substituting into the constraint:
   \[
   x^2 + 0^2 = 1 \implies x = \pm 1
   \]
   Then,
   \[
   f(1, 0) = 2(1)^2 + 3(0)^2 = 2
   \]
   \[
   f(-1, 0) = 2(-1)^2 + 3(0)^2 = 2
   \]

6. **Case 3: \( x \neq 0 \) and \( y \neq 0 \)**:
   Setting \( 2 = \lambda \) and \( 3 = \lambda \) gives a contradiction, so we ignore this case.

7. **Summary of critical points**:
   - \( f(0, 1) = 3 \)
   - \( f(0, -1) = 3 \)
   - \( f(1, 0) = 2 \)
   - \( f(-1, 0) = 2 \)

8. **Conclusion**:
   - **Maximum value**: \( 3 \) at points \( (0, 1) \) and \( (0, -1) \)
   - **Minimum value**: \( 2 \) at points \( (1, 0) \) and \( (-1, 0) \)

Would you like further details on any specific step, or do you have any questions? Here are some follow-up questions you might consider:

1. How does the method of Lagrange multipliers apply to other types of constraints?
2. Can you derive the gradients of the functions in a different context?
3. What are some real-world applications of optimization with constraints?
4. How would the solution change if the constraint were different?
5. What is the geometric interpretation of the constraint in this problem?

**Tip**: Always visualize the constraint and the function to better understand the optimization problem.

x^2 + y^2 = 1 and then f(x,y) = 2x^2 + 3y^2 do lagrange and find min max

Explore how to find the extrema of the function f(x,y) = 2x^2 + 3y^2 subject to the constraint x^2 + y^2 = 1 using Lagrange multipliers. This problem illustrates key optimization techniques in calculus, identifying maximum and minimum values.

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