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(A) The feasible region determined by the system of constraints $x + 3y \geq 3$,$x + y \geq 2$,and $x, y \geq 0$ is unbounded.The corner points of the

Vedclass · Accepted Answer

$7$

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(A) The feasible region determined by the system of constraints $x + 3y \geq 3$,$x + y \geq 2$,and $x, y \geq 0$ is unbounded.The corner points of the feasible region are $A(3, 0)$,$B(\frac{3}{2}, \frac{1}{2})$,and $C(0, 2)$.The values of $Z = 3x + 5y$ at these corner points are:      Corner Point    $Z = 3x + 5y$        $A(3, 0)$    $3(3) + 5(0) = 9$        $B(\frac{3}{2}, \frac{1}{2})$    $3(\frac{3}{2}) + 5(\frac{1}{2}) = \frac{9}{2} + \frac{5}{2} = \frac{14}{2} = 7$        $C(0, 2)$    $3(0) + 5(2) = 10$  Since the feasible region is unbounded,we check if $Z We draw the line $3x + 5y = 7$. It can be observed from the graph that the open half-plane $3x + 5y Therefore,the minimum value of $Z$ is $7$ at the point $(\frac{3}{2}, \frac{1}{2})$.

Solve the Linear Programming Problem graphically:
Minimise $Z = 3x + 5y$
subject to the constraints:
$x + 3y \geq 3$
$x + y \geq 2$
$x, y \geq 0$

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Solve the Linear Programming Problem graphically:Minimise $Z = 3x + 5y$subject to the constraints:$x + 3y \geq 3$$x + y \geq 2$$x, y \geq 0$