The solution set of the inequation 4x + 3y > 5 is

The corner points of the feasible region are $(0, 0)$(0,0), $(0, 40)$(0,40), $(20, 40)$(20,40), $(60, 20)$(60,20), $(60, 0)$(60,0).

The objective function is $z = 4x + 3y$z=4x+3y.

In linear programming, the maximum or minimum value of the objective function occurs at one of the corner points of the feasible region.

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At point $(0, 0)$(0,0):

$$z = 4(0) + 3(0)$$z=4(0)+3(0)

$$z = 0$$z=0

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At point $(0, 40)$(0,40):

$$z = 4(0) + 3(40)$$z=4(0)+3(40)

$$z = 0 + 120$$z=0+120

$$z = 120$$z=120

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At point $(20, 40)$(20,40):

$$z = 4(20) + 3(40)$$z=4(20)+3(40)

$$z = 80 + 120$$z=80+120

$$z = 200$$z=200

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At point $(60, 20)$(60,20):

$$z = 4(60) + 3(20)$$z=4(60)+3(20)

$$z = 240 + 60$$z=240+60

$$z = 300$$z=300

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At point $(60, 0)$(60,0):

$$z = 4(60) + 3(0)$$z=4(60)+3(0)

$$z = 240 + 0$$z=240+0

$$z = 240$$z=240

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Comparing all values:

At $(0, 0)$(0,0): $z = 0$z=0

At $(0, 40)$(0,40): $z = 120$z=120

At $(20, 40)$(20,40): $z = 200$z=200

At $(60, 20)$(60,20): $z = 300$z=300

At $(60, 0)$(60,0): $z = 240$z=240

The maximum value of $z = 300$z=300 occurs at point $(60, 20)$(60,20).