The coordinates of the corner points of the bounded feasible region are $(10, 0), (2, 4), (1, 5)$ and $(0, 8)$ . The maximum of objective function $z = 60 x + 10 y$ is

700

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800

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600

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110

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Solution

The correct option is C $600$
As the corner points are already given, substitute in $z = 60 x + 10 y$ to find the maximum of objective function.
For $(10, 0), z = 600 + 0 = 600$
For $(2, 4), z = 120 + 40 = 160$
For $(1, 5), z = 60 + 50 = 110$
For $(0, 8), z = 0 + 80 = 80$

Hence, $z_{max} = 600$

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The coordinates of the corner points of the bounded feasible region are (10,0), (2,4), (1,5) and (0,8). The maximum of objective function z=60x+10y is

The correct option is C 600As the corner points are already given, substitute in z=60x+10y to find the maximum of objective function. For (10,0), z=600+0=600 For (2,4), z=120+40=160 For (1,5), z=60+50=110 For (0,8), z=0+80=80 Hence, zmax=600

The coordinates of the corner points of the bounded feasible region are $(10, 0), (2, 4), (1, 5)$ and $(0, 8)$ . The maximum of objective function $z = 60 x + 10 y$ is

The correct option is C $600$
As the corner points are already given, substitute in $z = 60 x + 10 y$ to find the maximum of objective function.
For $(10, 0), z = 600 + 0 = 600$
For $(2, 4), z = 120 + 40 = 160$
For $(1, 5), z = 60 + 50 = 110$
For $(0, 8), z = 0 + 80 = 80$

Hence, $z_{max} = 600$