The minimum value of the objective function z = 2x + 10y for linear constraints x ≥ 0, y ≥ 0, x - y ≥ 0 , x - 5y ≤ - 5 , is

Question

The minimum value of the objective function z = 2x + 10y for linear constraints x ≥ 0, y ≥ 0, x - y ≥ 0 , x - 5y ≤ - 5 , is (A) 10 (B) 15 (C) 12 (D) 8

Tardigrade Admin · Accepted Answer

Required region is unbounded whose vertex is ( (5/4) , (5/4) ) . Hence the minimum value of objective function is = 2 × (5/4) + 10 × (5/4) = 15 .

Q. The minimum value of the objective function $z = 2 x + 10 y$ for linear constraints $x \geq 0, y \geq 0, x - y \geq 0, x - 5 y \leq - 5$ , is

10

15

12

8

Required region is unbounded whose vertex is $(\frac{5}{4}, \frac{5}{4}) .$
Hence the minimum value of objective function is
$= 2 \times \frac{5}{4} + 10 \times \frac{5}{4} = 15.$

Q. The minimum value of the objective function z=2x+10y for linear constraints x≥0,y≥0,x−y≥0,x−5y≤−5, is

10

15

12

8

Required region is unbounded whose vertex is (45​,45​). Hence the minimum value of objective function is =2×45​+10×45​=15.

Q. The minimum value of the objective function $z = 2 x + 10 y$ for linear constraints $x \geq 0, y \geq 0, x - y \geq 0, x - 5 y \leq - 5$ , is

Required region is unbounded whose vertex is $(\frac{5}{4}, \frac{5}{4}) .$
Hence the minimum value of objective function is
$= 2 \times \frac{5}{4} + 10 \times \frac{5}{4} = 15.$