To find the shortest and longest distances from the origin to the curve $x^2 + xy + y^2 = 4$ using the Lagrange multiplier method, follow these steps:

Step 1: Define the Objective Function

The squared Euclidean distance from the origin to a point $(x, y)$ is:

$f(x, y) = x^2 + y^2$

We aim to minimize and maximize $f(x, y)$ subject to the constraint:

$g(x, y) = x^2 + xy + y^2 - 4 = 0$

Step 2: Compute Gradients

The Lagrange function is:

$\mathcal{L}(x, y, \lambda) = x^2 + y^2 + \lambda (x^2 + xy + y^2 - 4)$

Taking partial derivatives, we get:

$\frac{\partial \mathcal{L}}{\partial x} = 2x + \lambda (2x + y) = 0$

$\frac{\partial \mathcal{L}}{\partial y} = 2y + \lambda (2y + x) = 0$

$\frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + xy + y^2 - 4 = 0$

Step 3: Solve the System

We solve:

1. $2x + \lambda (2x + y) = 0$
2. $2y + \lambda (2y + x) = 0$
3. $x^2 + xy + y^2 - 4 = 0$

From (1):

$\lambda = \frac{-2x}{2x + y}$

From (2):

$\lambda = \frac{-2y}{2y + x}$

Setting both expressions for $\lambda$ equal:

$\frac{-2x}{2x + y} = \frac{-2y}{2y + x}$

Cross multiplying:

$-2x(2y + x) = -2y(2x + y)$

$-4xy - 2x^2 = -4xy - 2y^2$

$-2x^2 = -2y^2$

$x^2 = y^2$

Thus, $x = \pm y$.

Step 4: Solve for x and y

Case 1: $x = y$

Substituting in $x^2 + xy + y^2 = 4$:

$x^2 + x^2 + x^2 = 4$

$3x^2 = 4$

$x^2 = \frac{4}{3}$

$x = \pm \frac{2}{\sqrt{3}}, \quad y = \pm \frac{2}{\sqrt{3}}$

Distance:

$d = \sqrt{x^2 + y^2} = \sqrt{\frac{4}{3} + \frac{4}{3}} = \sqrt{\frac{8}{3}} = \frac{2\sqrt{2}}{\sqrt{3}}$

Case 2: $x = -y$

Substituting in $x^2 + xy + y^2 = 4$:

$x^2 - x y + y^2 = 4$

$3x^2 = 4$

$x^2 = \frac{4}{3}$

$x = \pm \frac{2}{\sqrt{3}}, \quad y = \mp \frac{2}{\sqrt{3}}$

Distance:

$d = \sqrt{x^2 + y^2} = \sqrt{\frac{4}{3} + \frac{4}{3}} = \sqrt{\frac{8}{3}} = \frac{2\sqrt{2}}{\sqrt{3}}$

Since all distances are the same, both the shortest and longest distances are:

$\frac{2\sqrt{2}}{\sqrt{3}}$

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Final Answer:

• Shortest Distance = Longest Distance = $\frac{2\sqrt{2}}{\sqrt{3}}$

Would you like more clarification on any step?

Related Questions:

1. How does the Lagrange multiplier method work conceptually?
2. Can we solve this problem using another optimization technique?
3. What happens if the constraint curve is different?
4. How can we verify these results using parametric equations?
5. How do we generalize this approach to three dimensions?

Tip:

Always check whether the constrained function has only one extremum or multiple, as in some cases, both min and max can be different!