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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

  1. A 2
  2. B 5
  3. C 6
  4. D 3
Verified Solution

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.