question_answer

Maximize \[Z=4x+6y,\] subject to \[3x+2y\le 12,\]\[x+y\ge 4,x,y\ge 0,\]is

A) \[16\,\,at\,(4,0)\]

B) \[24\,\,at\,\,(0,4)\]

C) \[24\,\,at\,(6,0)\]

D) \[36\,\,at\,(0,6)\]

Correct Answer: D

Solution :

[d] we have, minimized \[Z=4x+6y\]

Subject to \[3x+2y\le 12,x+y\ge 4,x,y\ge 0\]

Let \[{{\ell }_{1}}:3x+2y=12\]

\[{{\ell }_{2}}:x+y=4\]

\[{{\ell }_{3}}:x=0\] and \[{{\ell }_{4}}:y=0\]

Shaded portion ABC is the feasible region, where A (4, 0), C (0, 4), B (0, 6).

Now maximize \[Z=4x+6y\]

\[Z\,\,at\,A\,(4,0)=4\,(4)+6\,(0)=16\]

\[Z\,\,at\,\,B\,(0,6)=4\,(0)+6\,(6)=36\]

\[Z\,\,at\,\,C\,(0,4)=4\,(0)+6\,(4)=24\]

Thus, Z is maximized at B (0, 6) and its maximum value is 36.