Question 1192848: Hi, I am having a lot of trouble with this question.
Let f(x, y) = x*y2 be a function with the definition set de (x, y) so that
2x^2+y^2=6
a) Explain how we can know that f has the definition set of maximum point and minimum point. Explain why f does not have stationary points.
b) Determine the largest and smallest value of f using the Lagranges method.
c) What is the answer to parts a) and b) if the definition set is 2x^2+y^2≤6 ?
d) Let h: R^2 → R be a function given by h(x,y)=4-x^2-y^2
i) Find the level curve of f at point P(1,−1). Determine ∇h(1,−1).
Draw the point P, as well as the level curve and the gradient to h at this
point.
ii) Calculate how fast h changes at point P in the direction
→ → →
v=−3i+4j
iii) In which direction from point P is this change greatest? (Hint:
the direction can be given as a vector).
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **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
*
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