Problems from Validity area of several simultaneos linear inequations

Development:

i) The straight lines associated with these inequations are (we will find them by taking the equality sign in the inequations and isolating the variable $y$): $$\begin{array}{rcl}f& :& 4x+5y=40\Rightarrow 5y=-4x+40\Rightarrow y=-{\displaystyle \frac{4}{5}}x+8\\ g& :& 2x+5y=30\Rightarrow 5y=-2x+30\Rightarrow y=-{\displaystyle \frac{2}{5}}x+6\\ h& :& x=0\\ i& :& y=0\end{array}$$

The last two straight lines are respectively the axis $y$ and the axis $x$.

Trying the point $(0,0)$ in the first two inequations we see that the regions of validity of these two restrictions are below the straight lines ($(0,0)$ is below the straight lines and satisfies the inequations). Concerning the other two restrictions, its validity regions are obvious: $x⩾0$ shows us that we can only have points to the right of the axis $y$ and the last restriction ($y⩾0$) shows us that its validity region is over the axis $x$. The validity area is bounded.

ii) We will firstly decide all the intersection points between the straight lines and then we will try which of them satisfy all the restrictions (these will be the apexes of the feasible region).

Intersection points:

• $f$ with $g$. They will cross at point $(x_0,y_0)$. We calculate as on other occasions: $$f(x_0)=g(x_0)\Rightarrow -{\displaystyle \frac{4}{5}}{x}_0+8=-{\displaystyle \frac{2}{5}}{x}_0+6\Rightarrow x_0=5$$ $$y_0=f(x_0)=g(x_0)=4\Rightarrow (x_0,y_0)=(5,4)$$ This point satisfies all the inequations, therefore it will be one of the apexes of the feasible region.

• $f$ with $h$. They will cross at point $(x_1,y_1)$. The straight line $h$ shows us that $x=0$, therefore the intersection point between two straight lines is: $$(0,f(0))=(0,8)\Rightarrow (x_1,y_1)=(0,8)$$ This point does not satisfy the inequation $2x+5y⩽30$, therefore this is not one of the apexes of the feasible region.

• $f$ with $i$: They will cross at point $(x_2,y_2)$. We calculate as on other occasions: $$f(x_2)=i(x_2)\Rightarrow -{\displaystyle \frac{4}{5}}{x}_2+8=0\Rightarrow x_2=10$$ $$y_2=f(x_2)=i(x_2)=0\Rightarrow (x_2,y_2)=(8,0)$$ This point satisfies all the inequations, therefore it will be one of the apexes of the feasible region.

• $g$ with $h$: They will cross at point $(x_3,y_3)$. The straight line $h$ shows us that $x=0$, therefore the intersection point between two straight lines is: $$(0,g(0))=(0,6)\Rightarrow (x_3,y_3)=(0,6)$$ This point satisfies all the inequations, therefore it will be one of the apexes of the feasible region.

• $g$ with $i$: They will cross at point $(x_4,x_4)$. Let's calculate these coordinates: $$g(x_4)=i(x_4)\Rightarrow -{\displaystyle \frac{2}{5}}{x}_2+6=0\Rightarrow x_4=15$$ $$y_4=g(x_4)=i(x_4)=0\Rightarrow (x_4,y_4)=(15,0)$$ This point does not satisfy the inequation $4x+5y⩽40$, therefore this is not one of the apexes of the feasible region.

• $h$ with $i$: They will cross at point $(x_5,x_5)$. The straight lines $h$ and $i$ say to us respectively that $x=0$ and $y=0$, therefore the intersection point will be where they the axes cut each other, that is to say in the origin: $(x_5,x_5)=(0,0)$. This point satisfies all the inequations, therefore it will be one of the apexes of the feasible region.

Solution:

i) The associated straight lines and regions of validity of the restrictions are:

• $f:y=-{\displaystyle \frac{4}{5}}x+8$ and its feasible region is below the straight line.
• $g:y=-{\displaystyle \frac{2}{5}}x+6$ and its feasible region is below the straight line.
• $h:x=0$ this straight line coincides with the axis $y$, and its feasible region is to the right of the above mentioned axis.
• $i:y=0$ this straight line coincides with the axis $x$, and its feasible region is over the above mentioned axis.

The common feasible area in all the inequations is bounded.

ii) The apexes of the common feasible area to all the restrictions are:

• $(x_0,y_0)=(5,4)$ (intersection point of $f$ with $g$).

• $(x_2,y_2)=(8,0)$ (intersection point of $f$ with $i$).

• $(x_3,y_3)=(0,6)$ (intersection point of $g$ with $h$).

• $(x_5,x_5)=(0,0)$ (intersection point of $h$ with $i$).

Not all the points, where the straight lines intersect, are in turn apexes of the feasible area. That's why we have had to verify what points all the restrictions satisfy simultaneously.

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