the constraints of a problem are listed below. what are the vertices of the feasible region?\n$2x + 3y \\geq…
the constraints of a problem are listed below. what are the vertices of the feasible region?\n$2x + 3y \\geq 12$\n$5x + 2y \\geq 15$\n$x \\geq 0$\n$y \\geq 0$\n\\(\\circ\\) \\((0, 0), (0, 4), \\left(\\frac{21}{11}, \\frac{30}{11}\\right), (3, 0)\\)\n\\(\\circ\\) \\((0, 0), \\left(0, \\frac{15}{2}\\right), \\left(\\frac{21}{11}, \\frac{30}{11}\\right), (6, 0)\\)\n\\(\\circ\\) \\((0, 4), \\left(\\frac{21}{11}, \\frac{30}{11}\\right), (3, 0)\\)\n\\(\\circ\\) \\(\\left(0, \\frac{15}{2}\\right), \\left(\\frac{21}{11}, \\frac{30}{11}\\right), (6, 0)\\)
Answer
Explanation:
Step1: Analyze (x = 0) case
For (x = 0), substitute into the inequalities (2x + 3y\geq12) and (5x + 2y\geq15).
- From (2(0)+3y\geq12), we get (3y\geq12\Rightarrow y\geq4).
- From (5(0)+2y\geq15), we get (2y\geq15\Rightarrow y\geq\frac{15}{2}=7.5). The intersection of (y\geq4) and (y\geq7.5) with (x = 0) and (y\geq0) is (y=\frac{15}{2}) (since (\frac{15}{2}>4)), so one vertex is ((0,\frac{15}{2})).
Step2: Analyze (y = 0) case
For (y = 0), substitute into the inequalities (2x + 3y\geq12) and (5x + 2y\geq15).
- From (2x+3(0)\geq12), we get (2x\geq12\Rightarrow x\geq6).
- From (5x + 2(0)\geq15), we get (5x\geq15\Rightarrow x\geq3). The intersection of (x\geq6) and (x\geq3) with (x\geq0) and (y = 0) is (x = 6), so one vertex is ((6,0)).
Step3: Find intersection of (2x + 3y=12) and (5x + 2y=15)
Solve the system (\begin{cases}2x+3y = 12\5x+2y=15\end{cases}) Multiply the first equation by (2): (4x + 6y=24) Multiply the second equation by (3): (15x+6y = 45) Subtract the first new equation from the second: (11x=21\Rightarrow x=\frac{21}{11}) Substitute (x = \frac{21}{11}) into (2x+3y=12): (2(\frac{21}{11})+3y=12\Rightarrow\frac{42}{11}+3y=12\Rightarrow3y=12-\frac{42}{11}=\frac{132 - 42}{11}=\frac{90}{11}\Rightarrow y=\frac{30}{11}) So the intersection point is ((\frac{21}{11},\frac{30}{11})).
Step4: Check ((0,0))
Check if ((0,0)) satisfies (2x + 3y\geq12): (2(0)+3(0)=0<12), so ((0,0)) is not in the feasible region. Also, check the other options, the vertices of the feasible region are ((0,\frac{15}{2}),(\frac{21}{11},\frac{30}{11}),(6,0))