Question

. Solve the following LP minimization problem.

        Min    3X + 2Y
        s.t.    5X + 3Y <= 30
                 3X + 4Y >= 36
                   Y >= 7
             X , Y >= 0

Group of answer choices

X = 0, Y= 9

The optimal value of the objective function is 5.

None of the other answers are correct.

The optimal value of the objective function is 7.

X = 1, Y = 3

Answer 1

min 3x + 2y 5X +34 < 30 3x + 4 V 236 Y 27 X, V 20 Constraint o 5x+34 L30 at Y=o, a point is 5x + 0 = 30 ..(x=6) so (60) is a point at t-o ,g point is 5(0) + 3y = 30 Y-10 then (6, 10) are another point so for the constraint 0 B (60) and (6, 10 are the points Constraint a 3x + 1 ) > 24 at 2=0, y = 36 = 9 1 (0,9) CS Scanned antih Yz0, = 36% = iz CamScanner (12,0)

for constraint ② points are (12,0) & (0,9) th the feasibu vagen * the feasible region is B solving constraint 0 A = (0,92, B=(0,10) and Ċ points gets by and constraint 5x + 3y = 30 3x + 4y = 36 multiply of fox & multiply with with 'y' for constraint '3 for second constraint 20% + 1/4 = 120 9x + 12 = 108 - 11x = 12 CS Scanned Withly cs Scanat, tot , y point is 5x +39 = 30 CamScanner

Volf C-5(1) Y = 8.18 8.18 Points objective function 7= 3x +24 minimum value is to be determined at point A=(0,9) Z = 310 + 2(9) Z .18 at point B = (0,10) . z = 310) + 2(10) at point c =12,8.18 del presidente - =3(3) + 2 (3.18) z = 19,63 so the minimum of objective function Scanned with CamScanner value - min {zo, Ze, Zc}

min { le, 20, 19.63) minimum objective = 18 - value " at FA at a points are (0,9). at points of (x=0 and V=9 Construction Constrant (ont Y >7 (0,0) (6,0) (12,07 x Constraint Constraint ③ 5x +39 530 3x +49 236 Constraint 5x+3y 230 'I' so the shaded area is towards I origin . from Constrain te ... 3x + 4 Y 736 '' so the shaded area is away - the origin Sorstan Y 2.7h it is also away from the origin CamScanner