The maximum value of 2x + y subject to 3x + 5 | vedclass

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(A) To find the maximum value of $Z = 2x + y$,we identify the corner points of the feasible region defined by the constraints $3x + 5y \leq 26$,$5x +

Vedclass · Accepted Answer

$12$

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(A) To find the maximum value of $Z = 2x + y$,we identify the corner points of the feasible region defined by the constraints $3x + 5y \leq 26$,$5x + 3y \leq 30$,$x \geq 0$,and $y \geq 0$.$1$. Find the intersection of the lines $3x + 5y = 26$ and $5x + 3y = 30$:Multiply the first equation by $5$ and the second by $3$:$15x + 25y = 130$$15x + 9y = 90$Subtracting the equations: $16y = 40 \implies y = \frac{40}{16} = 2.5$.Substitute $y = 2.5$ into $3x + 5(2.5) = 26 \implies 3x + 12.5 = 26 \implies 3x = 13.5 \implies x = 4.5$.So,point $B$ is $(4.5, 2.5)$.$2$. The corner points of the feasible region are $(0, 0)$,$(6, 0)$,$(4.5, 2.5)$,and $(0, 5.2)$.$3$. Evaluate $Z = 2x + y$ at each corner point:At $(0, 0): Z = 2(0) + 0 = 0$At $(6, 0): Z = 2(6) + 0 = 12$At $(4.5, 2.5): Z = 2(4.5) + 2.5 = 9 + 2.5 = 11.5$At $(0, 5.2): Z = 2(0) + 5.2 = 5.2$The maximum value is $12$ at the point $(6, 0)$.

The maximum value of $2x + y$ subject to $3x + 5y \leq 26$ and $5x + 3y \leq 30, x \geq 0, y \geq 0$ is

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