Question

The shaded region in the figure is the solution set of the inequations.

A. ${\text{5x + 4y}} \geqslant {\text{20,x}} \leqslant {\text{6,y}} \geqslant {\text{3,x}} \geqslant {\text{0,y}} \geqslant {\text{2}}$
B. ${\text{5x + 4y}} \geqslant {\text{20,x}} \leqslant {\text{6,y}} \geqslant {\text{3,x}} \geqslant {\text{0,y}} \geqslant {\text{2}}$
C. ${\text{5x + 4y}} \geqslant {\text{20,x}} \leqslant {\text{6,y}} \leqslant {\text{3,x}} \geqslant {\text{0,y}} \geqslant 0$
D. ${\text{5x + 4y}} \geqslant {\text{20,x}} \leqslant {\text{6,y}} \leqslant {\text{3,x}} \geqslant {\text{0,y}} \geqslant 2$

Answer 1

We’ll first find the equations of the line that are forming the area that has been shaded, after finding the equations of the line we’ll check for the points if that the area shaded is lying in the positive or negative i.e. above or below the particular line to create the inequalities we need.

Complete step by step Answer:
Now, to start with, we see,
We have our whole graph in the first quadrant,
${\text{x}} \geqslant {\text{0,y}} \geqslant 0$
Next, we can see the graph is enclosed below the line, ${\text{y = 3}}$
So, we can have, ${\text{y}} \leqslant {\text{3}}$,
Now, we can see the graph is enclosed on the left side of the line, ${\text{x = }}6$
So, we can have,${\text{x}} \leqslant 6$,
We know that equation of the line joining two points let say (a,b) and (c,d) is
$\Rightarrow \left( {y - b} \right) = \dfrac{{\left( {d - b} \right)}}{{\left( {c - a} \right)}}\left( {x - a} \right)$
Using the two-line form the equation of the line joining(0,5) and (4,0) is
$\Rightarrow \left( {y - 5} \right) = \dfrac{{\left( {0 - 5} \right)}}{{\left( {4 - 0} \right)}}\left( {x - 0} \right)$
$\Rightarrow \left( {y - 5} \right) = \dfrac{{ - 5}}{4}x$
$\Rightarrow 4y - 20 = - 5x$
$\Rightarrow 5x + 4y = 20$
Now, also, the graph is enclosed from the left by the line,${\text{5x + 4y = 20}}$,
Which is indicating away from the center with-respect-to the given graph.
So, the inequation should not satisfy the center.
Then we have the graph as, ${\text{5x + 4y}} \geqslant {\text{20}}$
If now we gather all of the equations together, we get,
${\text{5x + 4y}} \geqslant {\text{20,x}} \leqslant {\text{6,y}} \leqslant {\text{3,x}} \geqslant {\text{0,y}} \geqslant 0$

Hence option (C) will be the correct solution.

Note: Here the feasible region is made by four lines. Out of which two lines are parallel to the axes and one is the axis themselves. The other one is a given straight line with a negative slope with-respect-to the x-axis. We can also verify the inequalities by checking the point or the coordinated satisfying the inequalities.