The minimum value of the objective function $\mathrm{Z}=5 x+8 \mathrm{y}$, subject to $x+\mathrm{y} \geq 5$…
The minimum value of the objective function $\mathrm{Z}=5 x+8 \mathrm{y}$, subject to $x+\mathrm{y} \geq 5$
$x \leq 4, y \leq 2, x \geq 0, y \geq 0$ occur at the point
$(5,0)$
$(0,5)$
$(4,2)$
$(4,1)$
Solution
Feasible area is shaded.
Vertices of the feasible region are $A \equiv(4,1), B \equiv(4,2), C \equiv(3,2)$
$\therefore Z(A)=(5 \times 4)+(8 \times 1)=20+8=28$
$Z(B)=(5 \times 4)+(8 \times 2)=20+16=36$
$Z(C)=(5 \times 3)+(8 \times 2)=15+16=31$
Minima is at $(4,1)$