MHT CET · Maths · Linear Programming
The optimal solution of the L.P.P. Maximize : \(\mathrm{Z}=8 x+3\) y subject to the constraints \(x+y \leq 3,4 x+y \leq 6, x \geq 0, y \geq 0\) is
- A \(x=0, y=3\)
- B \(x=0, y=0\)
- C \(x=\frac{3}{2}, y=0\)
- D \(x=1, y=2\)
Answer & Solution
Correct Answer
(D) \(x=1, y=2\)
Step-by-step Solution
Detailed explanation
Here \(\mathrm{O} \equiv(0,0), \mathrm{A} \equiv\left(\frac{3}{2}, 0\right), \mathrm{C}=(0,3)\)
Point of intersection of given lines is \(\mathrm{B} \equiv(1,2)\)
\(Z=8 x+3 y\) and feasible region is shaded.
\(Z_{(0)}=0\)
\(Z_{(A)}=8\left(\frac{3}{2}\right)=12\)
\(Z_{(C)}=3(3)=9\)
\(Z_{(B)}=8(1)+3(2)=14\)
Point of intersection of given lines is \(\mathrm{B} \equiv(1,2)\)
\(Z=8 x+3 y\) and feasible region is shaded.
\(Z_{(0)}=0\)
\(Z_{(A)}=8\left(\frac{3}{2}\right)=12\)
\(Z_{(C)}=3(3)=9\)
\(Z_{(B)}=8(1)+3(2)=14\)
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