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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
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value of $\mathrm{Z}$ is
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Verified Answer
The correct answer is:
5
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.
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.
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