question_answer

The Maximum value of \[z=5x+3y\], subjected to the conditions\[3x+5y\le 15,5x+2y\le 10,x,y\ge 0\]is

A) \[\frac{235}{19}\]

B) \[\frac{325}{19}\]

C) \[\frac{523}{19}\]         

D) \[\frac{532}{19}\]

Correct Answer: A

Solution :

[a] Given, inequalities are \[3x+5y\le 15,\]\[4x+2y\le 10,x,y\ge 0\] Also, given \[z=5x+3y\] At point A (2, 0) \[z=5\times 2+0=10\] At Pont \[B\left( \frac{20}{19},\frac{45}{19} \right)\] \[z=\frac{5\times 20}{19}+\frac{3\times 45}{19}=\frac{235}{19}\] At point \[C(0,3)\] \[z=5(0)+3\times 3=9\] Hence, maximum value of z is\[\frac{235}{19}\].