# Math Help - Inequalities

1. ## Inequalities

Hi guys,

ok, lets hope you can help me out!

$x$ and $y$ must lie within or on the boundary of shaded region of my drawing

* What are the two constraints on the problem?

* Whats the max. value of the $f$ $3x + 6y$ within the shaded region?

Thankyou!

Hi guys,

ok, lets hope you can help me out!

$x$ and $y$ must lie within or on the boundary of shaded region of my drawing

* What are the two constraints on the problem?

* Whats the max. value of the $f$ $3x + 6y$ within the shaded region?

Thankyou!
If the points of the borderlines belong to the shaded region too then the shaded region is described by:

$y\leq-\frac52 x + 10~\wedge~y\leq -x+6~\wedge~x\geq 0~\wedge~y\geq 0$

If $f = 3x+6y~\implies~y=-\frac12 x+\frac16 f$ this line has to pass at least through one point of the shaded region. f gets a maximum if the line is translated as much up as possible.
Thus f is maximal if x = 0 and y = 6. In this case f = 36.

3. Originally Posted by earboth
If the points of the borderlines belong to the shaded region too then the shaded region is described by:

$y\leq-\frac52 x + 10~\wedge~y\leq -x+6~\wedge~x\geq 0~\wedge~y\geq 0$
So two options that represent the problem could be shown with what 2 equations?

Would these be two options that represent a constraint?

$y+\frac52 x \geq 10$ ?

$y\leq\frac52 x - 10$ ?

or

$y+\frac52 x \leq 10$ ?

$y\geq 10 + \frac25x$ ?

So two options that represent the problem could be shown with what 2 equations?

Would these be two options that represent a constraint?

$y+\frac52 x \geq 10$ ?

$y\leq\frac52 x - 10$ ?

or

$y+\frac52 x \leq 10$ ?

$y\geq 10 + \frac25x$ ?
I'm not sure if I understand you correctly...

Have a look at the graph which I posted with my previous reply:

$y\leq -x+6,\ if\ 0\leq x < \frac83~\wedge~\frac{10}3\leq y \leq 6$ .... and ..... $y\leq \frac52 x + 10 ,\ if\ \frac83 \leq x \leq 4 ~\wedge~0 \leq y \leq \frac83$

Since the line $y = -\frac12 x+\frac16 f$ has to pass through the highest point of the region build by the constraints, so that the value of f becomes a maximum, you have to look for this "highest" point. You find it in the first part of the constraints.