The point, at which the maximum value of $10 x+6 y$ subject to the constraints $x+y \leq 12$, $2 x+y \leq 20…

The point, at which the maximum value of $10 x+6 y$ subject to the constraints $x+y \leq 12$, $2 x+y \leq 20, x \geq 0, y \geq 0$ occurs, is
  1. $(10,0)$
  2. $(8,4)$
  3. $(0,12)$
  4. $(12,0)$

Solution

The feasible region lies on origin side of $x+y=12$ and $2 x+y=20$ and it is in the first quadrant.
The corner point of the feasible region are $\mathrm{O}(0,0), \mathrm{B}(10,0), \mathrm{C}(8,4), \mathrm{D}(0,12)$. At $\mathrm{O}(0,0) \mathrm{z}=10(0)+6(0)=0$ At B( 10,0$) z=10(10)+6(0)=100$ $\operatorname{AtC}(8,4) z=10(8)+6(4)=104$ At $\mathrm{D}(0,12) \mathrm{z}=10(0)+6(12)=72$ $\therefore \quad$ Maximum value of $z$ is 104 at it occurs at $C(8,4)$.

Asked in: MHT CET 2024 (02 May Shift 1)

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