CBSE 2023 · Region 3 · Set 2 · Q28 · 3 marks
Solve the following linear programming problem graphically : Maximize $\displaystyle \mathrm{z}=3 \mathrm{x}+9 \mathrm{y}$ subject to the constraints \[\begin{aligned} \mathrm{x}+\mathrm{y} & \geq 10 \\ \mathrm{x}+3 \mathrm{y} & \leq 60 \\ \mathrm{x} & \leq \mathrm{y} \\ \mathrm{x} \geq 0, \mathrm{y} & \geq 0 \end{aligned} \]
Marking-scheme solution
| Corner points | Value of $\displaystyle Z = 3x + 9y$ |
| $\displaystyle A(0, 20)$ | $\displaystyle 180 \rightarrow$ Maximum |
| $\displaystyle B(0, 10)$ | $\displaystyle 90$ |
| $\displaystyle C(5, 5)$ | $\displaystyle 60$ |
| $\displaystyle D(15, 15)$ | $\displaystyle 180 \rightarrow$ Maximum |
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.