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. |
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