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Minimise $\;z=2a+b\;$ subject to

$a\leqslant10$

$2a+5b\leqslant60$

$a+b\leqslant18$

$3a+b\leqslant44$

$a\geqslant0$, $\;\;b\geqslant0$

Usually I would plug the extreme points of the feasible region into the objective function but, for this question, would not it just be $0$ ?

Angelo
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TM1
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  • $(0,0)$ is also an extreme point. You did exactly what you said you would do, and you got the result you expected. What specifically is confusing you here? – JMoravitz Dec 28 '22 at 17:42
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    It may be worth keeping in mind that the question could have been written incorrectly and the intended question was to maximize the objective function instead. That would have required more thought and effort. – JMoravitz Dec 28 '22 at 17:44
  • @Jmoravitz I want to check if I understood it correctly because it seems to be too straightforward. Maybe there's a typo in the question – TM1 Dec 28 '22 at 17:44
  • @JMoravitz Agreed – TM1 Dec 28 '22 at 17:45
  • when in doubt draw a graphical representation on X-Y plane. You'd see feasible region is to the left (<=) of all the lines, bounded by the non-negative space. – Sutanu Majumdar Dec 28 '22 at 19:04
  • They had to have meant maximize unless this is to check your understanding. – John Douma Dec 28 '22 at 19:22

1 Answers1

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Using the Graphical Method, we can draw out the feasible region of the presented model as such:

enter image description here

with the feasible region of the model being the darkest region that is the intersection of all the constraints, and $a$ being the $x$-axis and $b$ being the $y$-axis.

It has been pointed out that the original problem may have been a maximization problem, so let’s take it as both interpretations starting with the minimization problem:

enter image description here

Depicting the objective function as the black line, we can see the most-minimum point to exist in the feasible region to satisfy the minimized model is the point where $a=0$ and $b=0$.

The maximization problem is a little more tricky, be we can see that if we slide the objective function’s line all the way to the right, we’ll reach a point that’s the highest possible value for the model like so:

enter image description here

and that point will correlate to intersection of the constraints $2a+5b\le60$ and $a+b\le18$. After solving for that intersection, we should find $a=10$ and $b=8$ such that $z=28$, which is the maximum value for the model.

Miss Mae
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