Find $\min \{ x+y: x+2y \ge 5, 4x+y\ge6\}$ Could anyone tell me what is the answer? Is it zero?
I drew all the lines: $x+y=0$ which intersect the 2nd line at $(2,-2)$.
and with the first line at $(-5,5)$.
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1$\begingroup$ You should graph it up. My guess is that it's obvious from a picture. $\endgroup$– Leonhard EulerJul 15, 2020 at 11:41
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$\begingroup$ @stuartstevenson, it is. Provided, of course, you understand the geometric interpretation of linear programming. $\endgroup$– Barry CipraJul 15, 2020 at 11:45
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$\begingroup$ You can also note that $7(x+y)=3(x+2y)+(4x+y)\geq 3\cdot 5+6=21$. Therefore, $x+y\geq 3$. Show that the inequality can become an equality. $\endgroup$– BatominovskiJul 15, 2020 at 11:45
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$\begingroup$ @BarryCipra And that you know what numbers are. $\endgroup$– Leonhard EulerJul 15, 2020 at 11:46
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$\begingroup$ @stuartstevenson, ah, no, just just that it's obvious what you need to do to find them. $\endgroup$– Barry CipraJul 15, 2020 at 11:53
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2 Answers
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I think this graph is sufficient.
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The constraints can be written (with $s:=x+y$)$$s\ge5-y,\\s\ge\frac{6+3y}4.$$
As one of the bounds is decreasing and the other growing, the optimum is achieved when they are equal, $$y=2,\\s=3.$$