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(A) First,we graph the feasible region of the system of linear inequalities $(2)$ to $(5)$.The feasible region $ABCD$ is shown in the figure. Note tha

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Minimum value is $60$ at $(5,5)$ and maximum value is $180$ at all points on the line segment joining $(15,15)$ and $(0,20)$.

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(A) First,we graph the feasible region of the system of linear inequalities $(2)$ to $(5)$.The feasible region $ABCD$ is shown in the figure. Note that the region is bounded.The coordinates of the corner points $A, B, C$ and $D$ are $(0,10), (5,5), (15,15)$ and $(0,20)$ respectively.Corner PointCorresponding value of $Z=3x+9y$$A(0,10)$$90$$B(5,5)$$60$ (Minimum)$C(15,15)$$180$ (Maximum)$D(0,20)$$180$ (Maximum)We find the minimum and maximum value of $Z$. From the table,the minimum value of $Z$ is $60$ at point $B(5,5)$.The maximum value of $Z$ on the feasible region occurs at the two corner points $C(15,15)$ and $D(0,20)$. Since the maximum value is the same at both points,any point on the line segment joining $C$ and $D$ will also give the maximum value of $180$.

Solve the following problem graphically:
Minimise and Maximise $Z=3x+9y$......$(1)$
subject to the constraints:
$x+3y \leq 60$.....$(2)$
$x+y \geq 10$......$(3)$
$x \leq y$.......$(4)$
$x \geq 0, y \geq 0$......$(5)$

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Solve the following problem graphically:Minimise and Maximise $Z=3x+9y$......$(1)$subject to the constraints:$x+3y \leq 60$.....$(2)$$x+y \geq 10$......$(3)$$x \leq y$.......$(4)$$x \geq 0, y \geq 0$......$(5)$