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The maximum value of Z=3x+4yZ = 3x + 4y subjected to the constraints 3x+7y≤213x + 7y \leq 21, 5x+2y≤105x + 2y \leq 10; x,y≥0x, y \geq 0 is :

Solution

✅ Correct Option: 4

Corners of feasible region: (0,0)(0,0), (2,0)(2,0), (0,3)(0,3), and intersection of 3x+7y=213x+7y=21 with 5x+2y=105x+2y=10 at (2829,7529)\left(\frac{28}{29}, \frac{75}{29}\right). At this point: z=8429+30029=38429z = \frac{84}{29} + \frac{300}{29} = \frac{384}{29}, the maximum.

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