question
If $x,y,z$ are real numbers such that $3x+y+2z \geq 3$ and $-x+2y+4z \geq 5$ find the smallest value of the real number $x+2y+4z$.
my idea
I forgot that we can't decrease to inequality so this is my non helpful idea:
Hope one of you can help me! Thank you!
Best Answer
The crucial point here is that in general, if you have a region $S$ delimited by two planes, the expression $x+2y+4z$ (or any non-zero linear combination of $x,y,z$) will range over the whole of $\Bbb R$ as $(x,y,z)$ ranges over $S$. In other words, there will be no smallest value.
The exception occurs when $x+2y+4z$ can be expressed as a linear combination $$x+2y+4z=A(3x+y+2z) + B(-x+2y+4z)$$ from which it would follow that $x+2y+4z$ must be at least $3A+5B$.
So can you find $A$ and $B$ so that the above equation holds? You need $3A-B=1$ (from the coefficient of $x$), $A+2B=2$ (from the coefficient of $y$), and $2A+4B=4$ (from the coefficient of $z$). In general such $A$ and $B$ won't exist, but you know that they must exist if this question is to have a valid solution.