Question 1192848
**a) Existence of Maximum and Minimum Points**

* **Closed and Bounded Set:** The constraint 2x² + y² = 6 defines an ellipse, which is a closed and bounded set in the plane. 
* **Continuous Function:** The function f(x, y) = x*y² is a polynomial, hence continuous everywhere, and therefore continuous on the ellipse.

**Extreme Value Theorem:** Since f is continuous on a closed and bounded set (the ellipse), the Extreme Value Theorem guarantees that f attains both a maximum and a minimum value on that set.

* **No Stationary Points:** 
    * ∇f(x, y) = (y², 2xy) 
    * ∇f(x, y) = 0 only at (0, 0) 
    * The point (0, 0) does not satisfy the constraint 2x² + y² = 6. 

**b) Lagrange Multipliers**

* **Lagrangian:** 
   L(x, y, λ) = x*y² - λ(2x² + y² - 6) 

* **Partial Derivatives:**
    * ∂L/∂x = y² - 4λx 
    * ∂L/∂y = 2xy - 2λy 
    * ∂L/∂λ = -(2x² + y² - 6)

* **System of Equations:**
    * y² - 4λx = 0 
    * 2xy - 2λy = 0 
    * 2x² + y² = 6

* **Solving the System:**
    * From the second equation: 
        * 2y(x - λ) = 0 
        * y = 0 or x = λ 
    * If y = 0, then from the constraint: 
        * 2x² = 6 
        * x = ±√3 
        * This gives us the points (±√3, 0) 
    * If x = λ, then from the first equation: 
        * y² - 4x² = 0 
        * y² = 4x² 
        * Substitute into the constraint: 
            * 2x² + 4x² = 6 
            * x = ±1 
        * If x = 1, then y² = 4, so y = ±2 
        * If x = -1, then y² = 4, so y = ±2 
        * This gives us the points (1, 2), (1, -2), (-1, 2), and (-1, -2)

* **Evaluate f at the Critical Points:**
    * f(√3, 0) = 0 
    * f(-√3, 0) = 0 
    * f(1, 2) = 4 
    * f(1, -2) = 4 
    * f(-1, 2) = -4 
    * f(-1, -2) = -4

* **Conclusion:** 
    * **Maximum Value:** 4 at (1, 2) and (1, -2) 
    * **Minimum Value:** -4 at (-1, 2) and (-1, -2)

**c) Definition Set: 2x² + y² ≤ 6**

* **Interior Points:** We already determined that there are no stationary points within the interior of the ellipse.
* **Boundary:** The analysis in part (b) already considered the boundary (2x² + y² = 6).

* **Conclusion:** The maximum and minimum values remain the same as in part (b) because the boundary points still provide the extrema.

**d) Function h(x, y) = 4 - x² - y²**

**i) Level Curve and Gradient at P(1, -1)**

* **Level Curve:** 
    * f(1, -1) = 1 
    * The level curve of f at P(1, -1) is the set of points (x, y) such that f(x, y) = 1, which is the curve x*y² = 1.

* **Gradient of h:** 
    * ∇h(x, y) = (-2x, -2y) 
    * ∇h(1, -1) = (-2, 2)

**ii) Directional Derivative**

* **Unit Vector in the Direction of v:** 
    * ||v|| = √((-3)² + 4²) = 5 
    *