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The maximum value of $z=2 x+3 y$ subject to constraints $3 x-3 y ≥ 0,2 x+2 y ≤ 12, x ≥ 0, y ≥ 0$ occurs at the point |
$(0,0)$ $(6,0)$ $(0,6)$ $(3,3)$ |
$(3,3)$ |
The correct answer is Option (4) - $(3,3)$ $z=2 x+3 y$ $3 x-3 y ≥ 0,2 x+2 y ≤ 12$ $⇒x-y≥ 0,⇒x+y≤6$ $x, y ≥ 0$
finding intersection point $x=y$ $⇒x+y=6$ $⇒x=y=3$
maximum occurs at $B(3,3)$ |
