Show that the minimum of $Z$ occurs at more than two points.
Minimise and Maximise $Z = x + 2y$
subject to $x + 2y \geq 100, 2x - y \leq 0, 2x + y \leq 200; x, y \geq 0$.
Minimise and Maximise $Z = x + 2y$
subject to $x + 2y \geq 100, 2x - y \leq 0, 2x + y \leq 200; x, y \geq 0$.
(N/A) The feasible region is determined by the constraints:
$x + 2y \geq 100, 2x - y \leq 0, 2x + y \leq 200, x \geq 0, y \geq 0$.
The corner points of the feasible region are $A(0, 50), B(20, 40),$ and $C(50, 100)$.
The values of $Z = x + 2y$ at these corner points are:
Since the minimum value of $Z$ is $100$ at both $A(0, 50)$ and $B(20, 40)$,the minimum value occurs at all points on the line segment joining $A$ and $B$. Thus,the minimum occurs at more than two points.
$x + 2y \geq 100, 2x - y \leq 0, 2x + y \leq 200, x \geq 0, y \geq 0$.
The corner points of the feasible region are $A(0, 50), B(20, 40),$ and $C(50, 100)$.
The values of $Z = x + 2y$ at these corner points are:
| Corner point | $Z = x + 2y$ |
| $A(0, 50)$ | $0 + 2(50) = 100$ (Minimum) |
| $B(20, 40)$ | $20 + 2(40) = 100$ (Minimum) |
| $C(50, 100)$ | $50 + 2(100) = 250$ |
Since the minimum value of $Z$ is $100$ at both $A(0, 50)$ and $B(20, 40)$,the minimum value occurs at all points on the line segment joining $A$ and $B$. Thus,the minimum occurs at more than two points.
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