CBSE Class 12 Mathematics 2025: question 27

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Step-by-step solution

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1. Graph. Draw the lines $x-y=0$ (through $(0,0)$ and $(4,4)$) and $x-2y=-2$ (through $(-2,0)$ and $(0,1)$). The region $x-y\ge 0$ lies on and below $y=x$. The region $x-2y\ge -2$ lies on and below $x-2y=-2$ (the origin satisfies it). Together with $x\ge 0,y\ge 0$, the feasible region lies above the $x$-axis, under $y=x$ from $O$ to $A$, and under $x-2y=-2$ beyond $A$. It is unbounded. The lines $y=x$ and $x-2y=-2$ meet where $x-2x=-2$, i.e. at $A(2,2)$. So the corner points are $O(0,0)$ and $A(2,2)$.

   1½ marks

2. Values of Z at the corner points

   Corner point	$Z=x+2y$
   $O(0,0)$	0
   $A(2,2)$	6

   1 mark

3. The largest value at a corner is 6, but the region is unbounded. Draw the open half-plane $x+2y>6$ (boundary line through $(6,0)$ and $(0,3)$). It has points in common with the feasible region, e.g. $(10,0)$ gives $Z=10>6$. Hence $Z$ has no maximum value.

   ½ mark

Answer: Corner points O(0,0) and A(2,2) give Z = 0 and 6. The feasible region is unbounded and $x+2y>6$ meets it, so Z has no maximum value.

Revise the chapter

• Chapter: Linear Programming (notes, formulas, where marks are lost and every past question)