To find the maximum and minimum distances from the origin to the curve $3x^2 + 4xy + 6y^2 = 140$, we can approach the problem as follows:

Step 1: Express the Distance Formula

The distance from the origin $(0,0)$ to any point $(x, y)$ on the curve is given by the Euclidean distance formula: $d = \sqrt{x^2 + y^2}$ To find the extrema of this distance, we need to maximize or minimize $d^2 = x^2 + y^2$, as minimizing/maximizing $d^2$ will give the same result for $d$.

Step 2: Set up a Lagrange Multiplier Problem

We are trying to find the extrema of $f(x, y) = x^2 + y^2$ subject to the constraint given by the curve $g(x, y) = 3x^2 + 4xy + 6y^2 - 140 = 0$.

This leads us to use the method of Lagrange multipliers. We introduce a multiplier $\lambda$ and set up the following system of equations:

$\nabla f = \lambda \nabla g$ Where:

• $\nabla f = (2x, 2y)$ is the gradient of $f(x, y) = x^2 + y^2$.
• $\nabla g = (6x + 4y, 4x + 12y)$ is the gradient of $g(x, y) = 3x^2 + 4xy + 6y^2 - 140$.

So, we have the following system of equations:

1. $2x = \lambda (6x + 4y)$
2. $2y = \lambda (4x + 12y)$
3. $3x^2 + 4xy + 6y^2 = 140$

Step 3: Solve the System

We will solve this system of equations to find $x$, $y$, and $\lambda$, which will give us the points on the curve where the distance is maximum and minimum.

Let's first divide the equations 1 and 2 by 2:

1. $x = \lambda (3x + 2y)$
2. $y = \lambda (2x + 6y)$

Now solve these equations systematically, and then substitute into the constraint $3x^2 + 4xy + 6y^2 = 140$.

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Related Questions:

1. What are Lagrange multipliers and when are they used?
2. How do we interpret the gradients $\nabla f$ and $\nabla g$ geometrically?
3. Can the method of Lagrange multipliers find saddle points?
4. What is the difference between minimizing and maximizing the distance function $d^2$?
5. How would you solve the system numerically if exact solutions are hard?

Tip:

In optimization problems involving curves, Lagrange multipliers are especially useful when direct substitution is challenging.