Question

graph the feasible region for the follow system of inequalities by drawing a polygon around the feasible region. click to set the corner points.

$$ \\left\\{ \
LATEXBLOCK0
\ ight. $$

Explanation:

Step1: Find the \(x\) and \(y\) - intercepts of \(x + 6y=30\)

• For \(x\)-intercept (\(y = 0\)): \(x=30\), so the point is \((30,0)\)
• For \(y\)-intercept (\(x = 0\)): \(6y=30\Rightarrow y = 5\), so the point is \((0,5)\)

Step2: Find the \(x\) and \(y\) - intercepts of \(4x + 3y=36\)

• For \(x\)-intercept (\(y = 0\)): \(4x=36\Rightarrow x = 9\), so the point is \((9,0)\)
• For \(y\)-intercept (\(x = 0\)): \(3y=36\Rightarrow y = 12\), so the point is \((0,12)\)

Step3: Solve the system \(

$$\begin{cases}x + 6y=30\\4x + 3y=36\end{cases}$$

\)
Multiply the second equation by \(2\): \(8x+6y = 72\)
Subtract the first equation \(x + 6y=30\) from it:
\((8x+6y)-(x + 6y)=72 - 30\)
\(8x+6y-x - 6y=42\)
\(7x=42\Rightarrow x = 6\)
Substitute \(x = 6\) into \(x+6y=30\): \(6+6y=30\Rightarrow6y=24\Rightarrow y = 4\)
The intersection point is \((6,4)\)

Step4: Consider the non - negativity constraints \(x\geq0\) and \(y\geq0\)

The corner points of the feasible region are \((0,0)\), \((0,5)\), \((6,4)\), \((9,0)\)

Answer:

The corner points of the feasible region (to be clicked on the graph) are \((0,0)\), \((0,5)\), \((6,4)\), \((9,0)\)