I'm having trouble with the following question:
Show $|ax + by| \leq (a^{2} + b^{2})^{1/2}$ if $x^{2} + y^{2} = 1$ by finding the maximum and minimum values of $f(x, y) = ax + by$ on the unit circle.
I'm learning multivariable calculus on my own, and I came across this question. I've been struggling for a couple of hours, and I'd really appreciate some help. I approach this problem using Lagrange Multipliers.
Here's what I've tried so far:
Let $f(x, y) = ax + by$ and let $g(x, y) = x^2 + y^2 – 1$. Then, by lagrange multiplier method, we have
$$a = \lambda(2x) $$
and
$$b = \lambda(2y),$$
from which we get
$$\lambda = \frac{a}{2x} = \frac{b}{2y}.$$
Then, I divided the two equations and I found $x = \frac{ay}{b}$, which I plugged into the constraint equation, but I got nowhere. Can someone please help me with this problem?
Best Answer
The equations that you get are:$$\left\{\begin{array}{l}a=2\lambda x\\b=2\lambda y\\x^2+y^2=1.\end{array}\right.$$Assuming that $a,b\neq0$, then you get that $\lambda\neq0$, that $x=\frac a{2\lambda}$, and that $y=\frac b{2\lambda}$. So, from the third equation you get that$$\left(\frac a{2\lambda}\right)^2+\left(\frac b{2\lambda}\right)^2=1.$$From this, you get two values for $\lambda$: $\lambda=\pm\frac{\sqrt{a^2+b^2}}2$. So, $(x,y)=\pm\left(\frac a{\sqrt{a^2+b^2}},\frac b{\sqrt{a^2+b^2}}\right)$.
Now, deal with the cases $a=0$ and $b=0$.