Show that the minimum of $Z$ occurs at more than two points.
Minimise and Maximise $Z = x + 2 y$
subject to $x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200; x, y \geq 0$ .

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Solution

Objective function : $Z = x + 2 y$

We have to minimize and maximize Z on constraints
$x + 2 y \geq 100$
$2 x - y \leq 0$
$2 x + y \leq 200$
$x \geq 0, y \geq 0$

After plotting all the constraints on coordinate plane, we get the feasible region as shown in the image.

Corner points Value of $Z = x + 2 y$
(0,200) 400 (Maximum)
(0,50) 100 (Minimum)
(20,40) 100 (Minimum)
(50,100) 250
Hence, maximum value of Z will be 400 and minimum value of Z will be 100.

Since, Minimum of Z will occurs for all the points of line $x + 2 y = 100$ lying in the feasible region.

Hence, there are more than 2 values where minimum of Z will occur.

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Minimize and miximize Z = x + 2y subject to constraints are x + 2y $\geq$ 100, 2x - y $\leq$ 0, 2x + y $\leq$ 200 and x, y $\geq$ 0.

Maximize:Z=x+2y
Subject to the constraints: $x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200, x \geq 0, y \geq 0$
Solve the L.P.P. graphically.

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