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Maximum value of $z=12 x+3 y$, subject to constraints $x \geq 0, y \geq 0, x+y \leq 5$ and $3 x+y \leq 9$ is
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The correct answer is:
36
Given, constraints are $x \geq 0, y \geq 0, x+y \leq 5$ and $3 x+y \leq 9$ and $z=12 x+3 y$
Here, feasible region is $O A B C O$.
At point $O(0,0), z=12(0)+3(0)=0$
At point $A(3,0), z=12(3)+3(0)=36$
At point $B(2,3), z=12(2)+3(3)=33$
At point $C(0,5), z=12(0)+3(5)=15$
Hence, maximum value of $z$ is 36 .
Here, feasible region is $O A B C O$.
At point $O(0,0), z=12(0)+3(0)=0$
At point $A(3,0), z=12(3)+3(0)=36$
At point $B(2,3), z=12(2)+3(3)=33$
At point $C(0,5), z=12(0)+3(5)=15$
Hence, maximum value of $z$ is 36 .
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