If $\mathrm{Z}=7 x+y$ subject to $5 x+y \geq 5, x+y \geq 3, x \geq 0, y \geq 0$, then minimum value of $\mathrm{Z}$ is

• A

  2
• Correct answer
  B

  5
• C

  6
• D

  3

Solution

Feasible area is shaded. Point of intersection of given lines is $\mathrm{P} \equiv\left(\frac{1}{2}, \frac{5}{2}\right)$ Co-ordinates of points are as follows : $C \equiv(3,0) ; P \equiv\left(\frac{1}{2}, \frac{5}{2}\right)$ and $B \equiv(0,5)$ We have $Z=7 x+y$ $Z_{(c)}=21+0=21$ $\therefore \quad Z_{(B)}=0+5=5$ $Z_{(P)}=\frac{7}{2}+\frac{5}{2}=6$ Thus minimum value is 5.

Hence, the correct answer is option: 'B'.