If $a,b$ are positive integers and $x^2+y^2\leq 1$ then find the maximum of $ax+by$ without differentiation.

Question

If $x^2+y^2\leq 1$ then maximum of $ax+by$

Here what I have done so far.

Let $ax+by=k$ . Thus $by=k-ax$.

So we can have that $$b^2x^2+(k-ax)^2 \leq b^2$$

$$b^2x^2+k^2-2akx +a^2x^2-b^2\leq 0 $$

By re-writing as a quadratic of $x$ ,

$$(a^2+b^2)x^2-2akx +k^2-b^2\leq 0 $$

Since $a^2+b^2$ is positive , above quadratic has a minimum. Thus it to be negeative it must have roots.

So $$(-2ak)^2-4(a^2+b^2)(k^2-b^2) \geq 0$$

$$a^2k^2-(a^2+b^2)k^2+(a^2+b^2)b^2 \geq 0$$ $$(a^2+b^2) \geq k^2$$

So maximum of $k$ is $\sqrt{a^2+b^2}$

Is this correct ?

If it is correct any shorter method ? Thanks in advance.

copper.hat · Answer 1 · 2017-09-12T05:30:25.440

3

Use Cauchy Буняко́вський Schwarz, $|\langle (a,b), (x,y) \rangle| \le \|(a,b)\| \|x,y\|$. Choosing $(x,y)= {1 \over \|(a,b)\|} (a,b)$ shows equality.

edited Sep 12 '17 at 05:30

answered Sep 12 '17 at 05:08

copper.hat

178,207

Thank you. But I do not know Cauchy Schwartz inequality. – Angelo Mark Sep 12 '17 at 05:09
The second "author" is Hermann Schwarz, not Schwartz :) – Martin R Sep 12 '17 at 05:10
@MartinR: Thanks for catching that. It is a mistake I repeat frequently, I am afraid to admit. – copper.hat Sep 12 '17 at 05:31
You are not alone: https://math.stackexchange.com/search?q=Cauchy+Schwartz :) – Martin R Sep 12 '17 at 05:33
@MartinR: Old habits are hard to break :-(. – copper.hat Sep 12 '17 at 05:34

score 3 · Accepted Answer · answered Sep 12 '17 at 05:15

3

Let $x=r \cos \theta$ and $y=r \sin \theta$ with $r \leq 1$ so that $x^2+y^2 \leq 1$. Then,

$$ax+by=r(a\cos \theta+b\sin \theta)$$

But one may show that we may write,

$$a \cos \theta+b \sin \theta=\sqrt{a^2+b^2} \cos (\theta+\phi)$$

For some $\phi$.

Hence $ax+by \leq (1)(\sqrt{a^2+b^2})$.

answered Sep 12 '17 at 05:15

Ahmed S. Attaalla

19,199

Thanks, your welcome :) – Ahmed S. Attaalla Sep 12 '17 at 05:17

neonpokharkar · Answer 3 · 2017-09-12T08:02:11.740

We can solve this by geometry,

$$x^2+y^2\le 1$$ is area bounded by circle with centre at origin and radius 1

&

$$ax+by=c$$ Is a line

We want to find max value of c,

Note: a and b are positive so slope is negative.

We want to find (x,y) such that it lies in circle and on line but also that line gives us maximum intersect at y-axis (which is $=\frac{c}{b}$)

If i draw the circle, and couple of lines In above picture we see that y-intersect is max when line is tangent to circle

In above picture we see that y-intersect is max when line is tangent to circle

Using basic coordinate knowledge, we can deduce the web of line such that it is tangent to circle

and then get c.

Mark Bennet · Answer 4 · 2017-09-12T07:24:30.447

You can consider $$(ax+by)^2+(bx-ay)^2=(a^2+b^2)(x^2+y^2)\le a^2+b^2$$ so that $$(ax+by)^2\le a^2+b^2-(bx-ay)^2$$

Equality is attained when $x^2+y^2=1$ and $bx=ay$

How did I get that second square? Well I wanted to work with the square of the target expression and to get a term in $x^2+y^2$, so I needed to swap $a$ and $b$ around so I had equivalent terms in $x$ and $y$. Then I wanted to eliminate the cross term, so I put the minus sign in to achieve that.

It is sometimes useful to pick out the cross term and eliminate the squares by doing something like $$(ax+by)^2-(ax-by)^2=4abxy$$ - for example in working with an area represented by the cross term.

If $a,b$ are positive integers and $x^2+y^2\leq 1$ then find the maximum of $ax+by$ without differentiation.

4 Answers4