The minimum of the objective function Z = 2x + 10y for linear constraints x – y ≥ 0, x – 5y ≤ – 5, x ≥ 0, y ≥ 0, is ___________.

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Solution

z = 2x + 10y
s.t x − y ≥ 0
x − 5y ≤ – 5, x ≥ 0, y ≥ 0
The equality constraints corresponding to given constraints are,
x − y = 0, x − 5y = −5, x = 0, y = 0
The feasible region is given by

Since at $(\frac{5}{4}, \frac{5}{4}),$ value of z = 2x + 10y = 2 $(\frac{5}{4}) + 10 (\frac{5}{4})$ = 15
∴ Minimum value of z = 2x + 10y is 15.

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