Q.Solve the L.P.P. graphically: Minimize: $z=5x+2y$, Subject to, $5x+y\ge 10$, $x+y\ge 6$, $x\ge 0, y\ge 0$

Graph the feasible region for the two $\ge$ constraints, find corner points, and evaluate $z$ at each.

Constraints: $5x+y\ge10$, $x+y\ge6$, $x\ge0,\ y\ge0$.

Boundary line 1: $5x+y=10$ — passes through $(2,0)$ and $(0,10)$.

Boundary line 2: $x+y=6$ — passes through $(6,0)$ and $(0,6)$.

Point of intersection of the two lines: Solve $5x+y=10$ and $x+y=6$ together. Subtracting: $4x=4\Rightarrow x=1$, then $y=6-1=5$. Intersection point $=(1,5)$.

Since both constraints are $\ge$-type, the feasible region is the (unbounded) region lying on or above both lines, in the first quadrant. Comparing which line is the binding (outer) boundary: at $x=0$, line 1 requires $y\ge10$ while line 2 requires $y\ge6$ — line 1 is tighter; at $x=6$, line 2 requires $y\ge0$ while line 1 is already satisfied — line 2 is tighter beyond $x=1$.

So the feasible region's boundary (the corner points relevant to the minimum) runs from $(0,10)$ down along $5x+y=10$ to $(1,5)$, then along $x+y=6$ to $(6,0)$.

Corner points: $(0,10)$, $(1,5)$, $(6,0)$.

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