CBSE 2026 · Region 2 · Set 2 · Q28 · 3 marks
Solve the following linear programming problem graphically: Maximize $\displaystyle \mathrm{Z}=8000 x+12000 \mathrm{y}$ Subject to constraints \[\begin{aligned} & 3 x+4 \mathrm{y} \leq 60 \\ & x+3 \mathrm{y} \leq 30 \\ & x \geq 0, \mathrm{y} \geq 0 \end{aligned} \]
Marking-scheme solution
| Corner Point | Value of Z = 8000x + 12000y |
| O$\displaystyle (0, 0)$ | $\displaystyle 0$ |
| A$\displaystyle (0, 10)$ | $\displaystyle 1,20,000$ |
| B$\displaystyle (12, 6)$ | $\displaystyle 1,68,000$ |
| C$\displaystyle (20, 0)$ | $\displaystyle 1,60,000$ |
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.