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Please solve RD Sharma class 12 chapter Linear Programming exercise 29.2 question 1 maths textbook solution

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Answer: The maximum value of Z is \frac{235}{19}  at the point \left(\frac{20}{19}, \frac{45}{19}\right)

Hint: Use application of Directive formula.

Given: Maximize: Z = 5x + 3y

Solution: -The possible region determined by the system of constraints, 3 x+5 y \leq 15,5 x+2 y \leq 10, x \geq 0, \text { and } y \geq 0  as follows:

The corner points of the feasible region are   \mathrm{O}(0,0), \mathrm{A}(2,0), \mathrm{B}(0,3),   and    \mathrm{C}\left(\frac{20}{19}, \frac{45}{19}\right). The value of Z at these corner points are as follows.

 

    Corner point     z=5x+3y
    O(0,0)     0
    A(2,0)     10
    B(0,3)     9
    C\left ( \frac{20}{19},\frac{45}{19} \right )     \frac{235}{19}

 

Therefore, the maximum value of Z is \frac{235}{19}  at the point \left(\frac{20}{19}, \frac{45}{19}\right)

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