# Lagrange multiplier question with unit circle constraint

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?

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 $$b=0$$.

• 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 $|ax + by|$? – user400359 Oct 24 '18 at 10:09
• 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. – José Carlos Santos Oct 24 '18 at 10:10
• 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? – user400359 Oct 24 '18 at 10:12
• I would use the fact that $\frac{a^2+b^2}{\sqrt{a^2+b^2}}=\sqrt{a^2+b^2}$. – José Carlos Santos Oct 24 '18 at 10:13

Hint: By Cauchy Schwarz we get $$(ax+by)^2\le (a^2+b^2)(x^2+y^2)$$

• 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. – user400359 Oct 24 '18 at 9:54
• You can use the Lagrange Multiplier method, but you have forgotten to differentiate with respect o $\lambda$$– Dr. Sonnhard Graubner Oct 24 '18 at 9:58 • Why would one differentiate with respect to$\lambda\$? – robjohn Oct 24 '18 at 10:23

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 $$\lambda$$.

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, $$2\lambda=\pm\sqrt{a^2+b^2}\tag3$$