Question

Let x+y=12
if A=x^2+y^2.
calculate the minimum value of A.
I have no idea!

Answer 1

Using the Method of La Grange:

Minimize A subject to \(x+y=12\)
$$
g(x,y,\lambda)=x^2+y^2+\lambda(x+y-12)\\
g_x=2x+\lambda=0\implies x=-\lambda/2\\
g_y=2y+\lambda=0\implies y=-\lambda/2\\
g_\lambda=x+y-12\implies\lambda=-12\\
x=6\\
y=6
$$
http://en.wikipedia.org/wiki/Lagrange_multiplier

Answer 2

what is the upside down y thing? I've never come across it ..

Answer 3

"\(\lambda\)" is called the La Grange Multiplier. Also, note that \(g_\lambda=0\) above.

Answer 4

\(\lambda\) is the Greek Letter Lambda. The way you make it in latex is by typing \lambda.

Answer 5

That's not what I learned though. the method we're supposed to use is supposed to involve differentiation

Answer 6

@yaysocks, the Method of La Grange Multipliers involves the partial differentials:
$$
\Large{
g_x=\cfrac{\partial g}{\partial x}\\
g_y=\cfrac{\partial g}{\partial y}\\
g_\lambda=\cfrac{\partial g}{\partial \lambda}
}
$$
This is similar to how you differentiate a single variable and set it equal to zero to find its maximum or minimum. Here we are doing the same, but using multiple variables \(x\) and \(y\). By introducing \(\lambda \) we can add the constraint equation, thereby creating 3 equations and 3 unknowns -- a solvable system of equations.

The contour lines of \(x+y =12\) and \(x^2+y^2\) touch when the tangent vectors of the contour lines are parallel. Since the gradient of a function is perpendicular to the contour lines, this is the same as saying that the gradients of \(x+y =12\) and \(x^2+y^2\) are parallel. These partials reflect this relationship.

Answer 7

@yaysocks, here is a validation of my solution:

http://www.wolframalpha.com/input/?i=minimize+x^2%2By^2%2Cx%2By%3D12