# Find $\min \{ x+y: x+2y \ge 5, 4x+y\ge6\}$ [closed]

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 $$(-5,5)$$.

• You should graph it up. My guess is that it's obvious from a picture. Jul 15, 2020 at 11:41
• @stuartstevenson, it is. Provided, of course, you understand the geometric interpretation of linear programming. Jul 15, 2020 at 11:45
• 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. Jul 15, 2020 at 11:45
• @BarryCipra And that you know what numbers are. Jul 15, 2020 at 11:46
• @stuartstevenson, ah, no, just just that it's obvious what you need to do to find them. Jul 15, 2020 at 11:53

## 2 Answers

I think this graph is sufficient.

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, $$y=2,\\s=3.$$