So I'm doing this problem, but I need help organizing the process in which it should be done.
The solution lists two ways to solve this problem, and I have questions regarding both of them:
1.) I understand a distance formula is being used for the Lagrangian, but why is it being added to the other part with the lambda? Are they not supposed to be set equal to each other? 2.) When using the Lagrange method, what does it mean to use the distance and constrain it (having trouble linking it back to gradient vectors and tangent lines and such.)? 3.) How is it that you'd go about using the second method using similar tangent lines? I was thinking of rearranging the initial equations so that x is a function of y then finding dy/dx, setting them equal to one another, and solving for the points in which they are the same.
No, they’re not supposed to be set equal to each other. The Lagrange multiplier method is based on the idea of finding points at which the two gradients are parallel. This is usually expressed as one gradient being a scalar multiple of the other. Observe that the length and orientation of the gradient of the constraint function here is arbitrary: you can multiply the constraint function by an arbitrary nonzero scalar, which will multiply its gradient by the same value, without changing the constraint.
You’re tasked with finding the extrema of distances between a set of points and a line; the function that expresses this distance is therefore the objective function. The set of points is constrained to lie on a certain ellipse, so that’s going to generate your constraint function.
Imagine moving the line toward the ellipse until it just touches it. The distance that you moved the line is then the shortest distance from the ellipse to the line and the intersection point is therefore the closest point to the original line. Now, keep moving the line in the same direction until it just touches the ellipse again: the new intersection point is the farthest point on the ellipse from the original line. Algebraically, this process can be expressed as finding the values of $k$ for which the line $2x+y=k$ and the ellipse have exactly one point in common. Solve for $x$ or $y$ in that linear equation, substitute into the equation of the ellipse and examine the discriminant to get a quadratic equation for $k$. Or, if you’re familiar with dual conics, you can use that to get a quadratic equation for $k$ directly.