The teacher hasn't uploaded a solution for this question yet.
First, we need to graph the inequalities to find the feasible region. The constraints are: $4x + y \ge 20$ $2x + 3y \ge 30$ $x \ge 0$ $y \ge 0$
Find the intersection points of the lines: $4x + y = 20$ and $2x + 3y = 30$ Multiply the first equation by 3: $12x + 3y = 60$ Subtract the second equation from the modified first equation: $(12x + 3y) - (2x + 3y) = 60 - 30$ $10x = 30$ $x = 3$ Substitute $x = 3$ into the first equation: $4(3) + y = 20$ $12 + y = 20$ $y = 8$ So, the intersection point is $(3, 8)$. The intersection of $4x+y=20$ with $x=0$ is $(0,20)$. The intersection of $4x+y=20$ with $y=0$ is $(5,0)$. The intersection of $2x+3y=30$ with $x=0$ is $(0,10)$. The intersection of $2x+3y=30$ with $y=0$ is $(15,0)$.
The corner points of the feasible region are $(0, 20)$, $(3, 8)$, and $(15, 0)$.
Evaluate $Z = 18x + 10y$ at each corner point: At $(0, 20)$: $Z = 18(0) + 10(20) = 0 + 200 = 200$ At $(3, 8)$: $Z = 18(3) + 10(8) = 54 + 80 = 134$ At $(15, 0)$: $Z = 18(15) + 10(0) = 270 + 0 = 270$
The minimum value of $Z$ is $134$ at the point $(3, 8)$.
Final Answer: 134
AI generated content. Review strictly for academic accuracy.