Solve the following linear programming problem graphically: gathered Minimise Z=20 subject to the cc 3 x+y ≥ 9 x+y ≥ 7 x+2 y ≥ 8 x, y ≥ 0 gathered

In the half-plane 2 x + y <8, there is no point in common with the feasible region. Hence, the minimum value of Z is 8, which is attained at x =1, y =6.

CBSE 2025 · Region 2 · Set 3 · Q28 · 3 marks

Solve the following linear programming problem graphically : \[\begin{gathered} \text { Minimise } \mathrm{Z}=20 \\ \text { subject to the cc } \\ 3 \mathrm{x}+\mathrm{y} \geq 9 \\ \mathrm{x}+\mathrm{y} \geq 7 \\ \mathrm{x}+2 \mathrm{y} \geq 8 \\ \mathrm{x}, \mathrm{y} \geq 0 \end{gathered} \]

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Corner point	Value of $\displaystyle \mathrm{Z}=2 \mathrm{x}+\mathrm{y}$
$\displaystyle \mathrm{A}(0,9)$	$\displaystyle 9$
$\displaystyle \mathrm{~B}(1,6)$	$\displaystyle 8$
$\displaystyle \mathrm{C}(6,1)$	$\displaystyle 13$
$\displaystyle \mathrm{D}(8,0)$	$\displaystyle 16$

Solve the following linear programming problem graphically : \[\begin{gathered} \text { Minimise } \mathrm{Z}=20 \\ \text { subject to the cc } \\ 3 \mathrm{x}+\mathrm{y} \geq 9 \\ \mathrm{x}+\mathrm{y} \geq 7 \\ \mathrm{x}+2 \mathrm{y} \geq 8 \\ \mathrm{x}, \mathrm{y} \geq 0 \end{gathered} \]

Marking-scheme solution

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