MHT CET · Maths · Linear Programming
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
- B 5
- C 6
- D 3
Answer & Solution
Correct Answer
(B) 5
Step-by-step Solution
Detailed explanation
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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