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(A) The feasible region is determined by the system of constraints $(2)$ to $(4)$.$1$. For the line $x + 2y = 10$,the intercepts are $(10, 0)$ and $(0

Vedclass · Accepted Answer

$2300$

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(A) The feasible region is determined by the system of constraints $(2)$ to $(4)$.$1$. For the line $x + 2y = 10$,the intercepts are $(10, 0)$ and $(0, 5)$.$2$. For the line $3x + 4y = 24$,the intercepts are $(8, 0)$ and $(0, 6)$.$3$. The intersection point of $x + 2y = 10$ and $3x + 4y = 24$ is found by solving the equations:Multiply the first by $2$: $2x + 4y = 20$.Subtracting this from $3x + 4y = 24$ gives $x = 4$.Substituting $x = 4$ into $x + 2y = 10$ gives $4 + 2y = 10$,so $2y = 6$,$y = 3$.The intersection point is $(4, 3)$.The corner points of the feasible region are $(0, 5)$,$(0, 6)$,and $(4, 3)$.						Corner Point			Value of $Z = 200x + 500y$							$(0, 5)$			$200(0) + 500(5) = 2500$							$(0, 6)$			$200(0) + 500(6) = 3000$							$(4, 3)$			$200(4) + 500(3) = 800 + 1500 = 2300$			The minimum value of $Z$ is $2300$ at point $(4, 3)$.

Solve the following linear programming problem graphically:
Minimise $Z = 200x + 500y$.......$(1)$
subject to the constraints:
$x + 2y \geqslant 10$.......$(2)$
$3x + 4y \leqslant 24$.....$(3)$
$x \geqslant 0, y \geqslant 0$......$(4)$

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Solve the following linear programming problem graphically:Minimise $Z = 200x + 500y$.......$(1)$subject to the constraints:$x + 2y \geqslant 10$.......$(2)$$3x + 4y \leqslant 24$.....$(3)$$x \geqslant 0, y \geqslant 0$......$(4)$