Question
Solve the following linear programming problem graphically:
Minimize: Z = 3x + 9y
When: $x+3 y \leq 60$
$\begin{aligned} x+y & \geq 10 \\ x & \leq y \\ x & \geq 0, y \geq 0\end{aligned}$

Answer


Image
Vertices of feasible region are
A(0, 20) B(15, 15) C(5,5) D(0, 10)
Corner pointsValue of Z = 1000x + 600y
(0,20)
(15, 15)
(5,5)
(0, 10)
180
180
$60 \rightarrow$ minimum
90
$\therefore$Z = 60 is minimum at x = 5, y = 5

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