Q17: Maximum and minimum of at three corner points
The corner points of the feasible region in the graphical representation of a linear programming problem are (2, 72), (15, 20) and (40, 15). If be the objective function, then :
(a) z is maximum at (2, 72), minimum at (15, 20)
(b) z is maximum at (15, 20), minimum at (40, 15)
(c) z is maximum at (40, 15), minimum at (15, 20)
(d) z is maximum at (40, 15), minimum at (2, 72)
Value of at each corner point:
Why?
This is the corner point method. If z has a maximum or a minimum value on the feasible region, it occurs at a corner point. So we only need z at the three corners.
Why?
Substitute each point into . For example, at :
Similarly , , and .
So z is maximum (855) at (40, 15) and minimum (450) at (15, 20).
Why?
The largest of the three values is the maximum and the smallest is the minimum.
Watch out: has the biggest coordinate, but has the smaller coefficient 9, so it does not give the maximum. Options (a) and (d) fall into this.
Answer: (c) z is maximum at (40, 15), minimum at (15, 20)
Chapter 12, Linear Programming: Corner point method.
Question type: Both maximum and minimum, and a tie along an edge
How to spot it: click to see
Type 3. Both maximum and minimum, and a tie along an edge
Spot it: "Minimise and maximise ". Often one rule, like , is a line through the origin.
Here: The three corners are given, so go straight to one table: . The largest, 855 at , is the maximum and the smallest, 450 at , the minimum.
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