CBSE 2024 · Region 1 · Set 1 · Q30 · 3 marks
Solve the following linear programming problem graphically : Maximise $\displaystyle \mathrm{z}=500 \mathrm{x}+300 \mathrm{y}$, subject to constraints \[\begin{aligned} & \mathrm{x}+2 \mathrm{y} \leq 12 \\ & 2 \mathrm{x}+\mathrm{y} \leq 12 \\ & 4 \mathrm{x}+5 \mathrm{y} \geq 20 \\ & \mathrm{x} \geq 0, \mathrm{y} \geq 0 \end{aligned} \]
Marking-scheme solution
Max $\displaystyle \mathrm{z}=500 \mathrm{x}+300 \mathrm{y}$
$\displaystyle \operatorname{Max} \mathrm{z}=3200$ at $\displaystyle \mathrm{x}=4, \mathrm{y}=4$
| Corner Point | Z |
| $\displaystyle \mathrm{A}(0,4)$ | $\displaystyle 1200$ |
| $\displaystyle \mathrm{~B}(0,6)$ | $\displaystyle 1800$ |
| $\displaystyle \mathrm{C}(4,4)$ | $\displaystyle 3200$ |
| $\displaystyle \mathrm{D}(6,0)$ | $\displaystyle 3000$ |
| $\displaystyle \mathrm{E}(5,0)$ | $\displaystyle 2500$ |
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.