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The corner points of the bounded feasible region determined by the system of linear inequalities are (0,0)(0,0), (4,0)(4,0), (2,4)(2,4) and (0,5)(0,5). If maximum value of z=ax+byz = ax + by, where a,b>0a,b > 0, occurs at both (2,4)(2,4) and (4,0)(4,0) then

Solution

✅ Correct Option: 1

The corner points of the bounded feasible region are (0,0)(0,0), (4,0)(4,0), (2,4)(2,4) and (0,5)(0,5).

The maximum value of z=ax+byz = ax + by, where a,b>0a,b > 0, occurs at both (2,4)(2,4) and (4,0)(4,0).

In linear programming, when the maximum occurs at two corner points, those points must give the same value of zz.


At point (2,4)(2,4):

z=a(2)+b(4)z = a(2) + b(4)

z=2a+4bz = 2a + 4b


At point (4,0)(4,0):

z=a(4)+b(0)z = a(4) + b(0)

z=4az = 4a


Since both points give the maximum value:

2a+4b=4a2a + 4b = 4a

4b=4a−2a4b = 4a - 2a

4b=2a4b = 2a

2b=a2b = a

Therefore, a=2ba = 2b.

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