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
- 900
- 1000
- 1250
- 1300
- 1520
The Correct Option is C
Approach Solution - 1
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 $z = 1250$.
Approach Solution -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.
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