Illustrate the following on graph paper and shade the region which satisfies all the three inequalities at the same time : \(- x + 5y \leq 10 ; 3x - 4y \leq...
Question 1 Report
Illustrate the following on graph paper and shade the region which satisfies all the three inequalities at the same time :
\(- x + 5y \leq 10 ; 3x - 4y \leq 8\) and \(x > -1\).
Answer Details
Rewrite the boundary equations as
\[ -x+5y=10\quad\Rightarrow\quad y=2+\frac{x}{5}, \]
\[ 3x-4y=8\quad\Rightarrow\quad y=-2+\frac{3x}{4}, \]
together with the vertical boundary \(x=-1\).
Use the following values to plot the two straight lines.
| For \(-x+5y=10\) | \(x=-4\) | \(x=0\) | \(x=4\) |
|---|---|---|---|
| \(y=2+\frac{x}{5}\) | \(1.2\) | \(2\) | \(2.8\) |
| For \(3x-4y=8\) | \(x=-4\) | \(x=0\) | \(x=4\) |
|---|---|---|---|
| \(y=-2+\frac{3x}{4}\) | \(-5\) | \(-2\) | \(1\) |
Plot the graph below. Draw the first two boundary lines solid, since their inequalities include equality. Draw \(x=-1\) as a broken line, since \(x>-1\) does not include the boundary.
Using \((0,0)\) as a test point:
\[0\leq10,\qquad 0\leq8,\qquad 0>-1.\]
Hence select the side containing \((0,0)\) for each inequality. Therefore the required region is
\[\boxed{\;x>-1,\qquad -2+\frac{3x}{4}\leq y\leq2+\frac{x}{5}\;}\]
The two sloping lines meet at
\[2+\frac{x}{5}=-2+\frac{3x}{4}\Rightarrow x=\frac{80}{11},\quad y=\frac{38}{11}.\]
Thus shade the triangular region between the two solid lines, to the right of the broken line \(x=-1\). The broken left-hand edge is not part of the region.
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