Validity area of several simultaneos linear inequations

Example

We have the following restrictions: $$\begin{array}{rcl}x+4& ⩾& 4\\ y& ⩽& 4\\ y& ⩾& x\end{array}$$

that they take as associated straight lines: $$\begin{array}{l}r(x):y=-x+4\\ s(x):y=4\\ t(x):y=x\end{array}$$

We can visualize these straight lines and determine the semiplanes where every inequation is satisfied separately.

For the straight line $r$:

Since the inequation is not satisfied at point $(0,0)$, $0+0⪈4$, we see that the area of validity of the inequation is the semiplane over the straight line.

For the straight line $s$:

Since the inequation is satisifed at point $(0,0)$, $0⩽4$, we see that the area of validity of the inequation is the semiplane below the straight line.

For the straight line $t$:

Since the inequation is satisifed at point $(0,1)$, $1⩾0$ we see that the area of validity of the inequation is the semiplane over the straight line.

As a whole we have:

The area where all the semiplanes coincide is the feasible region. We see that in this case it is also a question of a bounded area.

Now the calculation of the apexes of this area has to be done. To do so, it will be necessary to know the point where the straight lines cross. We will have to find three intersection points: the one of the straight line $r$ with $s$, the one of $r$ with $t$ and the one of $s$ with $t$.

Example

Back to the example.

Intersection point between $r$ and $s$:

We have to find the coordinates $x$ and $y$ of the intersection point. We will call the coordinates of this point $x_0$ and $y_0$.

To find the coordinate $x$ where the straight lines cross, $x_0$, we equal two functions $r(x)=-x+4$ and $s(x)=4$ at the point where the straight lines cross ($x_0$):

$$r(x_0)=s(x_0)\Rightarrow -x_0+4=4\Rightarrow x_0=0$$

Therefore two straight lines cross at $x_0=0$.

Determining the coordinate $y$ of the intersection point $y_0$ is simple, for it is the value that both functions $r(x)$ and $s(x)$ take in $x=x_0$.

$$y_0=r(x_0)=s(x_0)\Rightarrow y_0=r(0)=s(0)=-0+4=4$$

Therefore the intersection point between the straight lines $r$ and $s$ is: $(x_0=0,y_0=4)$.

Intersection point between $r$ and $t$:

We proceed just as in the previous case. We equal the functions $r(x)=-x+4$ and $t(x)=x$ at the point of intersection, that this time will take as its coordinates $(x_1,y_1)$.

$$r(x_1)=t(x_1)\Rightarrow -x_1+4=x_1\Rightarrow x_1=2$$

As in the previous case, the coordinate $y$ of the intersection point, $y_1$, is equal to the value that the functions $r(x)$ and $t(x)$ take at the intersection point:

$$y_1=r(x_1)=t(x_1)=-2+4=2$$

And so, the coordinates of the point of intersection are: $(x_1=2,y_1=2)$.

Intersection point between $s$ and $t$:

This intersection point will have as its coordinates $(x_2,y_2)$. First we determine the value of $x_2$ as in the previous cases, that is to say by equaling $s(x)=4$ and $t(x)=x$ at the intersection point $x_2$: $$s(x_2)=t(x_2)\Rightarrow 4=x_2$$

We decide the value of the coordinate $y$ of the point of intersection, $y_2$, as in the previous occasions: $$y_2=s(x_2)=t(x_2)=4$$

Therefore the intersection point between the straight lines $s(x)$ and $t(x)$ is the one that takes as its coordinates: $(x_2=4,y_2=4)$.

In short, the coordinates of the apexes of the feasible region are: $$(x_0,y_0)=(0,4)(x_1,y_1)=(2,2)(x_2,y_2)=(4,4)$$

Example

Considering the following inequations system: $$\begin{array}{rcl}x& ⩾& 0\\ y& ⩽& 4\\ y& ⩾& {\displaystyle \frac{x}{2}}\end{array}$$

We look first for the straight lines associated with every inequation and the areas of validity of each one:

• The first of them gives us a straight line parallel to the axis $y$ in $x=0$ and its validity region is the one that has $x$ greater than $0$ (towards the right of the axis $y$).
• The second one is a straight line parallel to the axis $x$, that meets the point $y=4$ and takes as a validity region the semiplane that it has below (where $y$ is less than $4$).
• The third straight line is $y={\displaystyle \frac{x}{2}}$ and its validity region is the one that is above the straight line (it is possible to verify this easily, seeing that the point $(0,1)$ satisfies the inequation: $1⩾$).

We will determine the apexes of the area of validity as the intersection points between the different straight lines.

• The straight line $x=0$ crosses with $y=4$ at point $(x_0=0,y_0=4)$.
• The straight line $x=0$ crosses with $y={\displaystyle \frac{x}{2}}$ at point $(x_1=0,y_1=0)$.
• The straight line $y=4$ crosses with $y={\displaystyle \frac{x}{2}}$ at point $(x_2=8,y_2=0)$.

Therefore, the apexes of the region of validity are: $$(x_0,y_0)=(0,4)(x_1,y_1)=(0,0)(x_2,y_2)=(8,0)$$

Example

Given the set of restrictions:

$$\begin{array}{rcl}x+3& ⩾& y\\ 8& ⩾& x+y\\ y& ⩾& x-3\\ x& ⩾& 0\\ y& ⩾& 0\end{array}$$

We look first for the straight lines associated with every restriction and the areas of validity of every inequation (checking it with a point in the inequations). The associatd straight lines are:

$$\begin{array}{l}f:y=3+x\\ g:y=-x+8\\ h:y=x-3\\ i:x=0\\ j:y=0\end{array}$$

Drawing the straight lines and the validity areas we can visualize the validity region.

If we cannot do the drawing we have an alternative. Having so many straight lines, normally we will have more intersection points between the straight lines than apexes in the validity area. For this reason, not all the points of intersection between the different straight lines will be apexes of the region of validity. To admit which ones are the apexes of the region of validity the following will be done:

• All the intersection points are calculated between the different straight lines.

• Those points of intersection where all the restrictions are fulfilled simultaneously will be the apexes of the area of validity (what can allow us to visualize it, if we have not done so before).

Thus, we are going to calculate all the intersection points between the different straight lines.

• $f$ with $g$ will cross at point $(x_0,y_0)$. We calculate the coordinates of the point as has already been done before: $$f(x_0)=g(x_0)\Rightarrow 3+x_0=-x_0+8\Rightarrow x_0={\displaystyle \frac{5}{2}}$$ And the coordinate $y$: $$y_0=f(x_0)=g(x_0)={\displaystyle \frac{11}{2}}$$ Therefore the coordinates of the point of intersection are: $(x_0,y_0)=({\displaystyle \frac{5}{2}},{\displaystyle \frac{11}{2}})$.

• $f$ with $h$ will cross at point $(x_1,y_1)$. We calculate the coordinates of the point as has already been done before: $$f(x_1)=h(x_1)\Rightarrow 3+x_1=x_1-3\Rightarrow 3=-3$$ We see that, when we try to find the coordinate x at the intersection point, an equation that is not satisfied appears. This means that two straight lines are in fact parallel (therefore they never cross).

• $f$ with $i$ will cross at point $(x_2,y_2)$. The straight line $f(x)=3+x$ will cross the straight line $x=0$ (straight line that coincides with the axis $y$) with point $(x=0,y=f(0))$. That is to say: $(x_2,y_2)=(0,3)$.

• $f$ with $j$ will cross at point $(x_3,y_3)$. We calculate the coordinates of the point as has already been done before: $$f(x_3)=j(x_3)\Rightarrow 3+x_3=0\Rightarrow x_3=-3$$ And the coordinate $y$, $$y_3=f(x_3)=j(x_3)=0$$ Therefore the coordinates of the point of intersection are: $(x_3,y_3)=(-3,0)$.

• $g$ with $h$ will cross at point $(x_4,y_4)$. We calculate the coordinates of the point as has already been done before: $$g(x_4)=h(x_4)\Rightarrow -x_4+8=x_4-3\Rightarrow x_4={\displaystyle \frac{11}{2}}$$ And the coordinate $y$: $$y_4=g(x_4)=h(x_4)={\displaystyle \frac{5}{2}}$$ Therefore the coordinates of the point of intersection are: $(x_4,y_4)=({\displaystyle \frac{11}{2}},{\displaystyle \frac{5}{2}})$.

• $g$ with $i$ will cross at point $(x_5,y_5)$. The straight line $i$ tells us $x=0$, therefore the intersection point will be: $(0,g(0))=(0,8)$. Therefore the coordinates of the point of intersection are: $(x_5,y_5)=(0,8)$.

• $g$ with $j$ will cross at point $(x_6,y_6)$. We calculate the coordinates of the point as has already been done before: $$g(x_6)=j(x_6)\Rightarrow -x_6+8=0\Rightarrow x_6=8$$ And the coordinate $y$: $$y_6=g(x_6)=j(x_6)=0$$ Therefore the coordinates of the point of intersection are: $(x_6,y_6)=(8,0)$.

• $h$ with $i$ will cross at point $(x_7,y_7)$. The straight line $i$ says to us that $x=0$, therefore the intersection point will be: $(0,h(0))=(0,-3)$. Therefore the coordinates of the point of intersection are: $(x_7,y_7)=(0,-3)$.

• $h$ with $j$ will cross at point $(x_8,y_8)$. We calculate the coordinates of the point as has already been done before: $$h(x_8)=j(x_8)\Rightarrow x_8-3=0\Rightarrow x_8=3$$ And the coordinate $y$: $$y_8=h(x_8)=j(x_8)=0$$ Therefore the coordinates of the point of intersection are: $(x_8,y_8)=(3,8)$.

• $i$ with $j$ will cross at point $(x_9,y_9)$. The straight line $i$ tells us that $x=0$, and $j$ that $y=0$, therefore the coordinates of the point of intersection are: $(x_9,y_9)=(0,0)$.

Determination of the apexes of the area of validity:

We have that nine intersection points between the straight lines. As we have said before, it has to be verified at what points all the inequations are satisfied, and these will be the apexes of the region of validity.

$$\begin{array}{l}(x_0,y_0)=({\displaystyle \frac{5}{2}},{\displaystyle \frac{11}{2}})\text{ all the inequations are satisfied.}\\ (x_2,y_2)=(0,3)\text{ all the inequations are satisfied.}\\ (x_3,y_3)=(-3,0)\text{ the inequation is not satisfied }x⩾0.\\ (x_4,y_4)=({\displaystyle \frac{11}{2}},{\displaystyle \frac{5}{2}})\text{ all the inequations are satisfied.}\\ (x_5,y_5)=(0,8)\text{ the inequation is not satisfied }x+3⩾y.\\ (x_6,y_6)=(8,0)\text{ the inequation is not satisfied }y⩾x-3.\\ (x_7,y_7)=(0,-3)\text{ the inequation is not satisfied }y⩾0.\\ (x_8,y_8)=(3,8)\text{ all the inequations are satisfied.}\\ (x_9,y_9)=(0,0)\text{ all the inequations are satisfied.}\end{array}$$

Therefore the apexes of the region of validity are: $$(x_0,y_0)=({\displaystyle \frac{5}{2}},{\displaystyle \frac{11}{2}})(x_2,y_2)=(0,3)(x_4,y_4)=({\displaystyle \frac{11}{2}},{\displaystyle \frac{5}{2}})$$ $$(x_8,y_8)=(3,8)(x_9,y_9)=(0,0)$$