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$ x ≥ 0$,

$ y ≥ 0$,

$2x + y < 8$

$x + 2y < 10$

Does x + y have a maximum value under the above conditions?

How I tried to do it:

I knew that x and y are positive numbers, and if trying to find the maximum value, they are most likely positive numbers.

I solve the system of inequalities 2x + y < 8, nd x + 2y < 10, to find possible values of x and y.

I get $ 2 < x$, and $y < 4$.

After plugging in those values for x and y respectively, I find that one of the inequalities gives an answer that does not satisfy the inequality.

$2x + y < 8$,

$2(2) + (4) = 8$

RHS must be greater than LHS, therefore that is not the case.

In order to satisfy the inequality, one number must be less than the other by 1. That means the maximum sum of the values x and y is equal to 5.

I have no idea if I'm right, and if this is the 'proper' way to do it so to speak, so guidance, other ideas, etc. would be really helpful. Tips on approaching minimum/maximum problems would be cool too. I'm just starting out in this kind of stuff.

Thank you!

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    $\begingroup$ get some graph paper and draw a really, really careful picture of the feasible region, meaning the region where all four given inequalities apply. $\endgroup$
    – Will Jagy
    Aug 22, 2014 at 1:56
  • $\begingroup$ Note that your objective function has no maximum subject to the constraints you have listed down. To tackle problems of this type you can use the graphical method. Look up linear programming via graphical method. $\endgroup$
    – Radz
    Aug 22, 2014 at 2:02
  • $\begingroup$ You say "In order to satisfy the inequality, one number must be less than the other by 1. That means the maximum sum of the values x and y is equal to 5." Does this mean you are looking for integers? If so, add that to the question. $\endgroup$ Aug 22, 2014 at 4:09

3 Answers 3

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enter image description here

In the picture $a$ is the line $2x+y=8$, $b$ is $x+2y=10$, and $c$ is $x+y=0$. Your problem is equivalent to finding the point in the pink polygon of the farthest distance to the line $c$. Can you see which point that is?

Also, note that your conditions are strict inequalities.

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  • $\begingroup$ Could we solve the system of equations for the lines that intersect that point, farthest from line c? $\endgroup$
    – user164403
    Aug 22, 2014 at 2:01
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    $\begingroup$ This would still not give a maximum; as you noticed, this will give you a point at which the inequalities are false. The system, as stated, does not have a maximum: no matter what point you choose inside the region, there is always another point where x+y is larger and slightly closer to (2,4). $\endgroup$ Aug 22, 2014 at 4:25
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The two latter inequalities imply $x+y<6$ so $x+y$ can not equal six. On the other hand, the point $x=2, y=4$ belongs to the boundary of the region. Therefore, $x+y$ being a continous function can be arbitrarily close to $6$, hence the answer is negative.

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Assuming that you are looking for integers:

First you need to remove the strict inequalities.

In order for $2x+y<8$ to be true, where x and y are integers, it must the case that $2x+y\leq7$

In order for $x+2y<10$ to be true, where x and y are integers, it must the case that $x+2y\leq9$

This system has maxima in the integers.

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