Lagrange multiplier question with unit circle constraint
By James Williams
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?
$\endgroup$5 Answers
$\begingroup$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 $b=0$.
$\endgroup$ 4 $\begingroup$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,$$ 2\lambda=\pm\sqrt{a^2+b^2}\tag3 $$
$\endgroup$ $\begingroup$Hint: By Cauchy Schwarz we get$$(ax+by)^2\le (a^2+b^2)(x^2+y^2)$$
$\endgroup$ 3 $\begingroup$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 $\lambda$.
$\endgroup$ 0 $\begingroup$You were actually well on your way to completing the proof. Inserting your result into the constraint equation yields
$$ x \ = \frac{ay}{b} \ \ \rightarrow \ \ \left( \frac{ay}{b} \right)^2 \ + \ y^2 \ \ = \ \ 1 \ \ \Rightarrow \ \ \left( \frac{a^2 \ + \ b^2}{b^2} \right) y^2 \ \ = \ \ 1 $$$$ y^2 \ \ = \ \ \frac{b^2}{a^2 \ + \ b^2} \ \ , \ \ x^2 \ \ = \ \ \left( \frac{a}{b} \right)^2 · \frac{b^2}{a^2 \ + \ b^2} \ \ = \ \ \frac{a^2}{a^2 \ + \ b^2} $$$$ \Rightarrow \ \ x \ = \ \pm \frac{|a|}{\sqrt{a^2 \ + \ b^2}} \ \ , \ \ y \ = \ \pm \frac{|b|}{\sqrt{a^2 \ + \ b^2}} \ \ , $$
as José Carlos Santos shows and which is presumably what you found.
The next question that arises is: what are we to do with this result? What you have found represents coordinates of tangent points to the unit circle for lines represented by the function $ \ ax \ + \ by \ $ . If we take, for instance, $ \ a > 0 \ , b > 0 \ $ (the other possible cases work similarly), then we have two lines with slope $ \ -\frac{a}{b} \ < \ 0 \ $ which contact the circle at $ \left( \frac{a}{\sqrt{a^2 \ + \ b^2}} \ , \ \frac{b}{\sqrt{a^2 \ + \ b^2}} \right) \ $ and $ \left( -\frac{a}{\sqrt{a^2 \ + \ b^2}} \ , \ -\frac{b}{\sqrt{a^2 \ + \ b^2}} \right) \ \ . $ These points lie on the lines
$$ ax \ + \ by \ \ = \ \ a·\frac{a}{\sqrt{a^2 \ + \ b^2}} \ + \ b·\frac{b}{\sqrt{a^2 \ + \ b^2}} \ \ = \ \ \frac{a^2 \ + \ b^2}{\sqrt{a^2 \ + \ b^2}} \ \ = \ \ \sqrt{a^2 \ + \ b^2} $$and $ \ \ ax \ + \ by \ \ = \ \ -\sqrt{a^2 \ + \ b^2} \ \ . $ [the geometric interpretation of robjohn's result]
All of the parallel lines passing through the circle are between these two lines, so we can conclude that for all points on the circle,
$$ -\sqrt{a^2 \ + \ b^2} \ \ \le \ \ ax \ + \ by \ \ \le \ \ \sqrt{a^2 \ + \ b^2} \ \ \ \text{or} \ \ \ | \ ax \ + \ by \ | \ \ \le \ \ \sqrt{a^2 \ + \ b^2} \ \ . $$
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