# Thread: some constrained extrema problems

1. ## some constrained extrema problems

ok so there a few problems giving me trouble.

1. Find the critical points of

cos^2 x + cos^2 y constrained to g(x,y) = x+ y = π/4

I used the larangian method and taking partials I managed to get sin 2x = sin 2y. Not sure how to solve that though for x in terms of y or vice versa.

2.

I'm trying to find the minimum of f(x,y) = x^2 + (y-2)^2 constrained to
g(x,y) = x^2 - y^2 = 1

taking the larangian method I got critical points as y = 1 and x = ± sqrt 2

I formed the hessian for f(x,y) and found it is positive defjnite and doesn't rely on the critical points at all. So in this case are both points minimums?

also since there is no maximum, would this contradict the EVT? I said yes because at the least there has to be one maximum on the interval, but I'm not sure if that's right or the right way to explain it.

2. ## Re: some constrained extrema problems

Why did you bother Mr. LaGrange with #1?

Notice how $y = \frac{\pi}{4} - x$

There is then nothing to suggest the function to be constrained isn't periodic.

The expression reduces, with no Frenchmen being awakened, to

$\cos^{2}(x) + \frac{1}{2}\cdot\sin(2x)+\frac{1}{2}$.

See if you can find the critical points of that.

3. ## Re: some constrained extrema problems

oh ok. Didn't know you could do that. The question requires lagrange though so how would I do it using that method. I got an equality

sin 2x = sin 2y. How do I solve that?

4. ## Re: some constrained extrema problems

Okay, let's head back to the revolution.

You got these:

1) $\lambda - 2\cdot\cos(x)\cdot\sin(x) = 0$

2) $\lambda - 2\cdot\cos(y)\cdot\sin(y) = 0$

3) $x - \frac{\pi}{4} + y = 0$

You used the first two to create $\sin(2x) = \sin(2y)$

You now throw in equation #3, no? This leads to a much simpler expression. Of course, it should be no surprise that it is conveniently related to the line x = y.