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(A) The feasible region is determined by the system of constraints:$3x + 5y \leq 15$,$5x + 2y \leq 10$,$x \geq 0$,and $y \geq 0$.The corner points of

Vedclass · Accepted Answer

$\frac{235}{19}$

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(A) The feasible region is determined by the system of constraints:$3x + 5y \leq 15$,$5x + 2y \leq 10$,$x \geq 0$,and $y \geq 0$.The corner points of the feasible region are $O(0, 0)$,$A(2, 0)$,and $B(0, 3)$. The intersection point $C$ of the lines $3x + 5y = 15$ and $5x + 2y = 10$ is found by solving the equations:Multiply the first by $2$ and the second by $5$:$6x + 10y = 30$$25x + 10y = 50$Subtracting the first from the second: $19x = 20 \implies x = \frac{20}{19}$.Substituting $x$ into $5x + 2y = 10$: $5(\frac{20}{19}) + 2y = 10 \implies \frac{100}{19} + 2y = 10 \implies 2y = 10 - \frac{100}{19} = \frac{90}{19} \implies y = \frac{45}{19}$.So,$C = (\frac{20}{19}, \frac{45}{19})$.The values of $Z$ at these corner points are:Corner Point$Z = 5x + 3y$$O(0, 0)$$0$$A(2, 0)$$10$$B(0, 3)$$9$$C(\frac{20}{19}, \frac{45}{19})$$5(\frac{20}{19}) + 3(\frac{45}{19}) = \frac{100 + 135}{19} = \frac{235}{19}$Thus,the maximum value of $Z$ is $\frac{235}{19}$ at the point $(\frac{20}{19}, \frac{45}{19})$.

Solve the Linear Programming Problem graphically:
Maximise $Z = 5x + 3y$
subject to the constraints:
$3x + 5y \leq 15$
$5x + 2y \leq 10$
$x \geq 0, y \geq 0$

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Solve the Linear Programming Problem graphically:Maximise $Z = 5x + 3y$subject to the constraints:$3x + 5y \leq 15$$5x + 2y \leq 10$$x \geq 0, y \geq 0$