CBSE 2023 · Region 2 · Set 3 · Q26 · 3 marks
Solve the following Linear Programming problem graphically : Maximize : $\displaystyle \mathrm{Z}=3 x+3.5 \mathrm{y}$ subject to constraints : $\displaystyle x+2 \mathrm{y} \geq 240$, $\displaystyle 3 x+1.5 \mathrm{y} \geq 270$, $\displaystyle 1.5 x+2 \mathrm{y} \leq 310$, $\displaystyle x \geq 0, \mathrm{y} \geq 0$.
Marking-scheme solution
Correct Graph
$\displaystyle \operatorname{Max}(\mathrm{Z})=595$ at $\displaystyle x=140 ; \mathrm{y}=50$
| Corner points | Value of Z |
| $\displaystyle \mathbf{A}(\mathbf{4 0 , 1 0 0})$ | $\displaystyle \mathbf{4 7 0}$ |
| $\displaystyle \mathbf{B}(\mathbf{1 4 0 , 5 0})$ | $\displaystyle \mathbf{5 9 5}($ Max $\displaystyle )$ |
| $\displaystyle \mathbf{C}(\mathbf{2 0 , 1 4 0})$ | $\displaystyle \mathbf{5 5 0}$ |
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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.