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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?

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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 $b=0$.

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  • $\begingroup$ Thanks. I don't understand how to relate this to $|ax + by| \leq \sqrt{a^2 + b^2}$ though. Do I just plug in $(x, y)$ back in for $|ax + by|$? $\endgroup$ – user400359 Oct 24 '18 at 10:09
  • $\begingroup$ Yes. It follows from what I did that the maximimum and the minimum are $\sqrt{a^2+b^2}$ and $-\sqrt{a^2+b^2}$ respectively. $\endgroup$ – José Carlos Santos Oct 24 '18 at 10:10
  • $\begingroup$ I plugged in your coordinate of $(x, y)$ in the positive case, and I found $|ax + by| = (a^2 + b^2)/(\sqrt{a^2 + b^2})$. What can I do from there? $\endgroup$ – user400359 Oct 24 '18 at 10:12
  • $\begingroup$ I would use the fact that $\frac{a^2+b^2}{\sqrt{a^2+b^2}}=\sqrt{a^2+b^2}$. $\endgroup$ – José Carlos Santos Oct 24 '18 at 10:13
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Hint: By Cauchy Schwarz we get $$(ax+by)^2\le (a^2+b^2)(x^2+y^2)$$

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  • $\begingroup$ Is there any way to do it without Cauchy Schwartz? This is part (a) to a question, and part (b) asks to prove the Cauchy Schwartz inequality, so I'm assuming that I'm not supposed to be using it for this initial part. $\endgroup$ – user400359 Oct 24 '18 at 9:54
  • $\begingroup$ You can use the Lagrange Multiplier method, but you have forgotten to differentiate with respect o $\lambda$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 24 '18 at 9:58
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    $\begingroup$ Why would one differentiate with respect to $\lambda$? $\endgroup$ – robjohn Oct 24 '18 at 10:23
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Instead of the equations you get, you can write

$$x=\frac{a}{2\lambda}\\ y=\frac{b}{2\lambda}$$

and plug that into $$x^2+y^2=1$$

which is the equation you get differentiating wrt $\lambda$.

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At the stationary point, when $a=2\lambda x$ and $b=2\lambda y$, we get the value to be $$ \begin{align} ax+by &=2\lambda\!\left(x^2+y^2\right)\\ &=2\lambda\tag1 \end{align} $$ Now to compute $\lambda$, $$ \begin{align} 1 &=x^2+y^2\\ &=\left(\frac a{2\lambda}\right)^2+\left(\frac b{2\lambda}\right)^2\tag2 \end{align} $$ Thus, $$ 2\lambda=\pm\sqrt{a^2+b^2}\tag3 $$

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