We are asked to find the minimum and maximum distance from the origin to the curve defined by the equation:

$3x^2 + 4xy + 6y^2 = 140$

Step 1: Expression for Distance

The distance $d$ from the origin to any point $(x, y)$ on the curve is given by the formula for Euclidean distance: $d = \sqrt{x^2 + y^2}$ To find the minimum and maximum distances, we need to optimize this function subject to the constraint $3x^2 + 4xy + 6y^2 = 140$.

Step 2: Lagrange Multipliers

We will use the method of Lagrange multipliers. The objective function to minimize or maximize is: $f(x, y) = x^2 + y^2$ The constraint is: $g(x, y) = 3x^2 + 4xy + 6y^2 - 140 = 0$ We introduce a Lagrange multiplier $\lambda$ and solve the system of equations obtained from the gradients of $f$ and $g$: $\nabla f = \lambda \nabla g$ This gives the system of equations:

1. $\frac{\partial}{\partial x}(x^2 + y^2) = \lambda \frac{\partial}{\partial x}(3x^2 + 4xy + 6y^2)$
2. $\frac{\partial}{\partial y}(x^2 + y^2) = \lambda \frac{\partial}{\partial y}(3x^2 + 4xy + 6y^2)$

Step 3: Compute Gradients

• $\nabla f(x, y) = (2x, 2y)$
• $\nabla g(x, y) = (6x + 4y, 4x + 12y)$

So the system of equations becomes: 2x = \lambda (6x + 4y) \tag{1} 2y = \lambda (4x + 12y) \tag{2}

Step 4: Solve the System

We now solve equations (1) and (2) along with the constraint equation $3x^2 + 4xy + 6y^2 = 140$.

Case 1: $x = 0$

If $x = 0$, from equation (1), we get: $2y = \lambda(12y)$ For $y \neq 0$, this implies $\lambda = \frac{1}{6}$. Substituting $x = 0$ into the constraint: $6y^2 = 140 \implies y^2 = \frac{140}{6} = \frac{70}{3}$ So: $y = \pm \sqrt{\frac{70}{3}}$ The distance is: $d = \sqrt{0^2 + \left(\sqrt{\frac{70}{3}}\right)^2} = \sqrt{\frac{70}{3}} = \frac{\sqrt{210}}{3}$

Case 2: $y = 0$

If $y = 0$, from equation (2), we get: $2x = \lambda (6x)$ For $x \neq 0$, this implies $\lambda = \frac{1}{3}$. Substituting $y = 0$ into the constraint: $3x^2 = 140 \implies x^2 = \frac{140}{3}$ So: $x = \pm \sqrt{\frac{140}{3}}$ The distance is: $d = \sqrt{\left(\sqrt{\frac{140}{3}}\right)^2 + 0^2} = \sqrt{\frac{140}{3}} = \frac{\sqrt{420}}{3}$

Step 5: Analyze Results

The possible distances are $\frac{\sqrt{210}}{3}$ and $\frac{\sqrt{420}}{3}$. Since $\sqrt{420} > \sqrt{210}$, the maximum distance is $\frac{\sqrt{420}}{3}$ and the minimum distance is $\frac{\sqrt{210}}{3}$.

Final Answer:

• Minimum distance: $\frac{\sqrt{210}}{3}$
• Maximum distance: $\frac{\sqrt{420}}{3}$

Would you like more details or have any questions?

Here are five related questions to explore further:

1. How do Lagrange multipliers work in multi-variable optimization?
2. What other methods could be used to find extremum points on constrained curves?
3. How can we solve systems of nonlinear equations numerically?
4. What are the geometric interpretations of Lagrange multipliers?
5. How do the shapes of different conic sections affect distances from the origin?

Tip: When using Lagrange multipliers, always check if the constraint is active by substituting back into the original equation.