Question

Maximam value of $:Z=10x+25$ y Subject to : $x\le 3,y\le 3,x+y\le 5,x\ge 0,y\ge 0$ at:

• (A) 95
• (B) 90
• (C) 85
• (D) 80

Answer 1

The Correct Option is A

Explanation:
Given:$Z=10x+25y$Subject to: $x\le 3,y\le 3,x+y\le 5,x\ge 0,y\ge 0$First we draw the lines AB, CD and EF whose equations are x = 3, y = 3 and x + y = 5 respectively.LineEquationPoint on theX-axisPoint on theY-axisSignRegionABx = 3A(3,0)-origin side of line ABCDy = 3-D(0,3)origin side of line CDEFx + y = 5E(5,0)F(0,5)origin side of line EF

The feasible region is OAPQDO which is shaded in the figure.The vertices of the feasible region are O (0,0), A (3, 0), P, Q and D (0, 3)P is the point of intersection of the lines x + y = 5 and x = 3Substituting x = 3 in x + y = 5, we get,3+ y=5 y = 2 P= (3, 2)Q is the point of intersection of the lines x + y = 5 and y = 3Substituting y = 3 in x + y = 5, we get,x + 3 = 5x = 2Q = (2,3)The values of the objective function z = 10x + 25y at these vertices areZ(O)  =10(0)+ 25(0)= 0Z(A)  =10(3) + 25(0) = 30Z(P)  =10(3) + 25(2) = 30 + 50 = 80Z(Q)  =10(2) + 25(3) = 20 + 75 = 95Z(D)  =10(0) + 25(3) =75Z has max imumvalue 95, when x = 2 and y = 3.Hence, the correct option is (A).