The maximum value of $Z=3x+4y $ subjected to the constraints $3x+7y ≤21, 5x+2y ≤10; x, y ≥ 0$ is :
Answer & explanation
Correct answer: option 4
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),
$3(\frac{28}{29})+7y=21$
$\frac{84}{29}+7y=21$
$⇒y=21-\frac{84}{29}=\frac{525}{29}$
$⇒y=\frac{525}{29×7}=\frac{75}{29}$
$Z=3x+2y$
$→(x,y)=\left(\frac{28}{29},\frac{75}{29}\right)$
$Z=3×\frac{28}{29}+4×\frac{75}{29}$
$=\frac{384}{29}$