The maximum value of $Z=3x+4y $ subjected to the constraints $3x+7y ≤21, 5x+2y ≤10; x, y ≥ 0$ is :

$\frac{384}{29}$

The correct answer is Option (4) → $\frac{384}{29}$

$3x+7y=21$    ...(1)

$5x+2y=10$    ...(2)

-2 × Eq. (1) + 7 × Eq. (2), we get

$35x+14y-6x-14y=70-42$

$29x=28$

$x=\frac{28}{29}$

Putting this value in eq. (1),

$3(\frac{28}{29})+7y=21$

$\frac{84}{29}+7y=21$

$⇒y=21-\frac{84}{29}=\frac{525}{29}$

$⇒y=\frac{525}{29×7}=\frac{75}{29}$

$Z=3x+2y$

$→(x,y)=\left(\frac{28}{29},\frac{75}{29}\right)$

$Z=3×\frac{28}{29}+4×\frac{75}{29}$

$=\frac{384}{29}$