Choose the correct alternative : The maximum value of z = 5x + 3y. subject to the constraints

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Z = 5x + 3y The inequalities are 3x + 5y $\leq$ 15, 5x + 2y $\leq$ 10 Consider lines L1 and L2 where L1 : 3x + 5y = 15, 5x + 2y = 10 For line L1, Plot A (0, 3) and B (5, 0) For line L2, plot P(0, 5) and Q(2, 0) Solving both line, we get x = (20)/(19), y = (45)/(19) The coordinates of the origin O (0, 0) satisfies both the inequalities. $\therefore$ The required region is on the origin side of both the lines L1 and L2. As x $\geq$ 0, y $\geq$ 0; the feasible region is in the 1st quadrant. OQRAO is the required feasible region. At O (0, 0), Z = 0 At Q (2, 0), Z = 5(2) + 0 = 10 At R `($\frac{20}{19}$, $\frac{45}{19}$) , z = 5($\frac{20}{19}$) + 3($\frac{45}{19}$) = (235)/(19)`. At A (0, 3), Z = 0 + 3(3) = 9 The maximum value of Z is (235)/(19) and it occurs at point R($\frac{20}{19}$, $\frac{45}{19}$).

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