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function: 4x-y+2 and constraint is 2x^2 +y^2 =1

The problem asks to find the maximum and minimum of the function \( f(x, y) = 4x - y + 2 \) subject to the constraint \( 2x^2 + y^2 = 1 \).

This is a constrained optimization problem, and we can solve it using the **Lagrange multiplier method**.

### Steps:

1. **Define the Lagrange multiplier function**:
   \[
   \mathcal{L}(x, y, \lambda) = f(x, y) - \lambda (g(x, y) - 1)
   \]
   Here, \( g(x, y) = 2x^2 + y^2 \) represents the constraint, and we set it equal to 1.

   So, the Lagrange function becomes:
   \[
   \mathcal{L}(x, y, \lambda) = (4x - y + 2) - \lambda(2x^2 + y^2 - 1)
   \]

2. **Find the partial derivatives**:
   We take the partial derivatives of \( \mathcal{L}(x, y, \lambda) \) with respect to \( x \), \( y \), and \( \lambda \).

   \[
   \frac{\partial \mathcal{L}}{\partial x} = 4 - \lambda (4x) = 0
   \]
   \[
   \frac{\partial \mathcal{L}}{\partial y} = -1 - \lambda(2y) = 0
   \]
   \[
   \frac{\partial \mathcal{L}}{\partial \lambda} = -(2x^2 + y^2 - 1) = 0
   \]

3. **Solve the system of equations**:
   From \( \frac{\partial \mathcal{L}}{\partial x} = 0 \):
   \[
   4 = 4\lambda x \quad \Rightarrow \quad \lambda = \frac{1}{x}
   \]
   
   From \( \frac{\partial \mathcal{L}}{\partial y} = 0 \):
   \[
   -1 = -2\lambda y \quad \Rightarrow \quad \lambda = \frac{1}{2y}
   \]

   Now, equate the two expressions for \( \lambda \):
   \[
   \frac{1}{x} = \frac{1}{2y}
   \quad \Rightarrow \quad x = 2y
   \]

4. **Substitute into the constraint** \( 2x^2 + y^2 = 1 \):
   \[
   2(2y)^2 + y^2 = 1
   \]
   \[
   2(4y^2) + y^2 = 1
   \]
   \[
   8y^2 + y^2 = 1 \quad \Rightarrow \quad 9y^2 = 1 \quad \Rightarrow \quad y^2 = \frac{1}{9} \quad \Rightarrow \quad y = \pm \frac{1}{3}
   \]

5. **Find corresponding values for \( x \)**:
   Since \( x = 2y \), when \( y = \frac{1}{3} \), \( x = \frac{2}{3} \), and when \( y = -\frac{1}{3} \), \( x = -\frac{2}{3} \).

6. **Evaluate the function** at the critical points:
   - For \( \left(\frac{2}{3}, \frac{1}{3}\right) \):
     \[
     f\left(\frac{2}{3}, \frac{1}{3}\right) = 4\left(\frac{2}{3}\right) - \frac{1}{3} + 2 = \frac{8}{3} - \frac{1}{3} + 2 = \frac{10}{3} + 2 = \frac{16}{3}
     \]
   - For \( \left(-\frac{2}{3}, -\frac{1}{3}\right) \):
     \[
     f\left(-\frac{2}{3}, -\frac{1}{3}\right) = 4\left(-\frac{2}{3}\right) - \left(-\frac{1}{3}\right) + 2 = -\frac{8}{3} + \frac{1}{3} + 2 = -\frac{7}{3} + 2 = -\frac{1}{3}
     \]

### Conclusion:
- The maximum value of the function is \( \frac{16}{3} \) at \( \left(\frac{2}{3}, \frac{1}{3}\right) \).
- The minimum value of the function is \( -\frac{1}{3} \) at \( \left(-\frac{2}{3}, -\frac{1}{3}\right) \).

Would you like to go over any details, or do you have any questions?

### Related Questions:
1. What is the geometric interpretation of Lagrange multipliers?
2. Can this method be applied to functions with more variables and constraints?
3. How would the solution change if the constraint equation were different?
4. What happens if there is no critical point in a constrained optimization problem?
5. Can we solve this problem graphically using contour plots?

### Tip:
Always check the second derivative or Hessian matrix for further confirmation of maximum or minimum values in optimization problems.

Find the maximum and minimum of the function f(x, y) = 4x - y + 2 subject to the constraint 2x^2 + y^2 = 1.

Learn how to find the maximum and minimum of the function f(x, y) = 4x - y + 2 using the Lagrange multiplier method, subject to the constraint 2x^2 + y^2 = 1. This guide provides a step-by-step solution suitable for undergraduate calculus students.

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