MHT CET2019Morning ShiftMathematicsArea Under CurvesActual
If z = a x + b y ; a , b > 0 subject to x ≤ 2 , y ≤ 2 , x + y ≥ 3 , x ≥ 0 , y ≥ 0 has minimum value at 2,1 only, then…
Options
- Aa > b
- Ba = b
- Ca < b
- Da = 1 + b
Correct answer
C. a < b
Step-by-step solution
We have, z = a x + b y , a , b > 0 Subject to constraints x ≤ 2 , y ≤ 2 , x + y ≥ 3 x , y ≥ 0 On taking given constraints as equation, we get the following graph Here, ABCA is the required feasible region whose corner points are A 2,1 , B 1,2 and C 2,2 . Since, It is given that z = a x + b y ; a , b > 0 has minimum value at 2,1 ∴ Value of z at 2,1 < value of z at 1,2 ⇒ 2 a + b < a + 2 b ⇒ a < b