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Show that the minimum of Z occurs at more than two points.
Minimize and Maximize Z = 5x + 10y subject to \(x + 2y \leqslant 120,x + y \geqslant 60\)\(x - 2y \geqslant 0,x,y \geqslant 0\).
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Show that the minimum of Z occurs at more than two points.
Minimize and Maximize Z = 5x + 10y subject to \(x + 2y \leqslant 120,x + y \geqslant 60\)\(x - 2y \geqslant 0,x,y \geqslant 0\).

Answer

Consider \(x + 2y \leqslant 120\)
Let x + 2y = 120
\( \Rightarrow \frac{x}{{120}} + \frac{y}{{60}} = 1\)
The half plane containing(0, 0) is the required half plane as (0, 0) makes \(x + 2y \leqslant 120\), true.
x03060
y604530
Answer for Show that the minimum of Z occurs at more than two points.Minimize and Maximize Z = 5x + 10y subject to x + 2y leqslant 120,x + y geqslant 60, x - 2y geqslant 0,x,y geqslant 0.
Again \(x + y \geqslant 60\)
Let x + y = 60
Also the half plane containing (0, 0) does not make \(x + y \geqslant 60\) true.
Therefore, the required half plane does not contain (0, 0).
Again \(x - 2y \geqslant 0\)
Let x - 2y = 0 \( \Rightarrow x = 2y\)
Let test point be (30, 0).
x03060
y01530
\( \Rightarrow x - 2y \geqslant 0 \Rightarrow 30 - 2 \times 0 \geqslant 0\) It is true.
Therefore, the half plane contains (30, 0).
The region CFEKC represents the feasible region.
At C (60, 0) Z \( = 5 \times 60 = 300\)
At F (120, 0) \(Z = 5 \times 120 = 600\)
At E (60, 30)\(Z = 5 \times 60 + 10 \times 30 = 600\)
At K (40, 20) \(Z = 5 \times 40 + 10 \times 20 = 400\)
Hence, minimum Z = 300 at x = 60, y = 0 and maximum Z = 600 at x = 120, y = 0 or x = 60, y = 30.
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