Question

The feasible region for a L.P.P. is shown in the figure below. Let z = 50x+15y be the objective function, then the maximum value of z is

• A. 900
• B. 1000
• C. 1250
• D. 1300
• E. 1520

Answer 1

The Correct Option is C

To find the maximum value of $z = 50x + 15y$, we need to evaluate the objective function at the vertices of the feasible region. From the graph, the coordinates of the vertices are:
A(20, 0)
B(10, 50)
C(0, 60)
We calculate $z$ at each of these points: For point A(20, 0): \[ z = 50(20) + 15(0) = 1000 \] For point B(10, 50): \[ z = 50(10) + 15(50) = 500 + 750 = 1250 \] For point C(0, 60): \[ z = 50(0) + 15(60) = 900 \] The maximum value of $z$ occurs at point B(10, 50) with $z = 1250$. 

Answer 2

Finding the Maximum Value of z = 50x + 15y

To determine the maximum value of z = 50x + 15y, we need to evaluate the objective function at the vertices of the feasible region. The coordinates of the vertices are:

• A(20, 0) 
• B(10, 50)
• C(0, 60)

We calculate the value of z at each of these points:

At point A(20, 0):
z = 50(20) + 15(0) = 1000

At point B(10, 50):
z = 50(10) + 15(50) = 500 + 750 = 1250

At point C(0, 60):
z = 50(0) + 15(60) = 900

The maximum value of z occurs at point B(10, 50) where z = 1250.