Consider an LPP: Maximise Z = 50x + 15y subjected to constraints x + y ≤ 60, 5x + y ≤ 100, x, y ≥ 0. If the maximum value of Z occurs at x = α and y…

In any LPP, the maximum or minimum value occurs at a corner point of the feasible region.

The corner points are:

Origin: $(0, 0)$(0,0)

Where $5x + y = 100$5x+y=100 meets the x-axis, set $y = 0$y=0:

$$5x = 100$$5x=100

$$x = 20$$x=20

Point: $(20, 0)$(20,0)

Where $x + y = 60$x+y=60 meets the y-axis, set $x = 0$x=0:

$$y = 60$$y=60

Point: $(0, 60)$(0,60)

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Intersection of the two lines:

$x + y = 60$x+y=60 ... (1)

$5x + y = 100$5x+y=100 ... (2)

Subtracting (1) from (2):

$$(5x + y) - (x + y) = 100 - 60$$(5x+y)−(x+y)=100−60

$$4x = 40$$4x=40

$$x = 10$$x=10

Substituting $x = 10$x=10 in equation (1):

$$10 + y = 60$$10+y=60

$$y = 50$$y=50

Point: $(10, 50)$(10,50)

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Evaluating $Z = 50x + 15y$Z=50x+15y at each corner point:

At $(0, 0)$(0,0): $Z = 50(0) + 15(0) = 0$Z=50(0)+15(0)=0

At $(20, 0)$(20,0): $Z = 50(20) + 15(0) = 1000$Z=50(20)+15(0)=1000

At $(0, 60)$(0,60): $Z = 50(0) + 15(60) = 900$Z=50(0)+15(60)=900

At $(10, 50)$(10,50): $Z = 50(10) + 15(50) = 500 + 750 = 1250$Z=50(10)+15(50)=500+750=1250

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The maximum value of $Z$Z is $1250$1250 at point $(10, 50)$(10,50).

Therefore, $\alpha = 10$α=10 and $\beta = 50$β=50

$$\alpha + \beta = 10 + 50 = 60$$α+β=10+50=60