# Ngô Quốc Anh

## February 2, 2010

### A question about how estimate a linear combination over a region

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 18:50

There is a question saying that how to estimate $x+y$ provided

$\displaystyle 1<2x+y<3$

and

$\displaystyle 2.

We may follow the following argument. The idea is to estimate $x$ and $y$ and the we just combine all together. In order to estimate $x$, you need to kill $y$, for example, multiple the first equation by $-\frac{1}{2}$

$\displaystyle -3<-\frac{x}{2}-y<-1$

and then add the second equation to the above equation. What we have is

$\displaystyle 2<\frac{3x}{2}<2$

Thus

$\displaystyle -\frac{4}{3}.

Similarly, one gets

$\displaystyle \frac{1}{3}.

Therefore, from above we obtain

$\displaystyle \boxed{-1.

However, if you add the given equations, you will get

$\displaystyle 3<3x+3y<9$.

Thus $\boxed{1. So what’s happening?

In order to see why, we need to figure out the domain of points $(x,y)$.

In the above picture, the blue region is the domain of $(x,y)$. Two blue lines are $x+y=-1$ and $x+y=5$. Thus, one can see that the interval $(-1,5)$ is too large. For e.g., there is no point $(x,y)$ in the blue region such that $(x,y)$ approaches these two blue lines. In other words, one can see that $(0,0)$ lies outside the region but the condition $-1 still holds.

OK, so now let consider the second case.

From the above picture, one can see that two blue lines (with equations $x+y=1$ and $x+y=3$) touch the blue region, therefore, one can find point $(x,y)$ in the blue region such that $x+y$ can achieve arbitrary value between $1$ and $3$. Therefore this is the correct estimate.

In general, if you want to deal with $\alpha x + \beta y$. Then for given $\alpha, \beta$ you have a family of lines $\alpha x + \beta y = \star$. This family depends on the value $\star$. Then the lower bound and upper bound for $\alpha x + \beta y$ are two numbers so that the corresponding blue lines touch the blue region.

This is very theoretical. If you want to find these numbers explicitly, we need to go back to the correct way solving our particular example mentioned above. What I am trying to say is to find two numbers called $a, b$ such that

$\displaystyle a(2x+y)+b(x+2y)=\alpha x + \beta y$.

Clearly, this is a system of linear equations, one can easy find $a=\frac{2\alpha-\beta}{3}$ and $b=\frac{2\beta-\alpha}{3}$.

Thus

$\displaystyle\underbrace {1 \times \frac{{2\alpha- \beta }}{3} + 2 \times \frac{{2\beta- \alpha }}{3}}_\beta< \alpha x + \beta y < \underbrace {3 \times \frac{{2\alpha- \beta }}{3} + 6 \times \frac{{2\beta- \alpha }}{3}}_{3\beta }.$