CBSE 2026 · Region 4 · Set 1 · Q29 · 3 marks
Solve the following Linear Programming Problem graphically: Maximise $\displaystyle \mathrm{Z}=200 \mathrm{x}+120 \mathrm{y}$ subject to the constraints \[\begin{array}{r} \mathrm{x}+\mathrm{y} \leq 300 \\ 3 \mathrm{x}+\mathrm{y} \leq 600 \\ \mathrm{x}-\mathrm{y} \geq-100 \\ \mathrm{x}, \mathrm{y} \geq 0 \end{array} \]
Marking-scheme solution
| Corner point | Value of Z = 200x + 120y |
| A $\displaystyle (100, 200)$ | $\displaystyle 44000$ |
| B $\displaystyle (150, 150)$ | $\displaystyle 48000$ |
| C $\displaystyle (0, 100)$ | $\displaystyle 12000$ |
| D $\displaystyle (0, 0)$ | $\displaystyle 0$ |
| E $\displaystyle (200, 0)$ | $\displaystyle 40000$ |
Linear ProgrammingLinear Programming Problem and its Mathematical FormulationApplynumericmedium
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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.