Question

Maximam value of :Z=10x+25 y Subject to : x≤3,y≤3,x+y≤5,x≥0,y≥0 at:

• A. (A) 95
• B. (B) 90
• C. (C) 85
• D. (D) 80

Answer 1

Answer: (A)

(A) 95

Answer 2

Explanation:
Given:Z=10x+25ySubject to: x≤3,y≤3,x+y≤5,x≥0,y≥0First 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 EFThe 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).