The maximum value of $z$ for the linear programing problem maximize $z = x + y$ subject to the constraints $x + 4y \leq 8, 2x + 3y \leq 12, 3x + y \leq 9, x \geq 0, y \geq 0$ is:

The objective is to maximize $z = x + y$ subject to:

$x + 4y \leq 8$

$2x + 3y \leq 12$

$3x + y \leq 9$

$x \geq 0,\; y \geq 0$

In linear programming, the maximum of $z$ always occurs at a corner point of the feasible region.

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Corner point $A$: Origin

$A = (0,\; 0)$

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Corner point $B$: $x$-axis intercept $(y = 0)$

From constraint 1: $x \leq 8$

From constraint 2: $x \leq 6$

From constraint 3: $x \leq 3$ ← most restrictive

$B = (3,\; 0)$

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Corner point $C$: $y$-axis intercept $(x = 0)$

From constraint 1: $y \leq 2$ ← most restrictive

From constraint 2: $y \leq 4$

From constraint 3: $y \leq 9$

$C = (0,\; 2)$

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Corner point $D$: Intersection of $x + 4y = 8$ and $3x + y = 9$

$x + 4y = 8 \quad \cdots(i)$

$3x + y = 9 \quad \cdots(ii)$

From $(ii)$: $y = 9 - 3x$

$x + 4(9 - 3x) = 8$

$x + 36 - 12x = 8$

$-11x = -28$

$x = \dfrac{28}{11}$

$y = 9 - 3\!\left(\dfrac{28}{11}\right)$

$= \dfrac{99 - 84}{11}$

$= \dfrac{15}{11}$

Checking constraint 2: $\;2\!\left(\dfrac{28}{11}\right) + 3\!\left(\dfrac{15}{11}\right) = \dfrac{56 + 45}{11} = \dfrac{101}{11} \approx 9.18 \leq 12$ ✅

$D = \left(\dfrac{28}{11},\; \dfrac{15}{11}\right)$

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The other intersections are not feasible:

Intersection of constraints 2 and 3 gives $\left(\dfrac{15}{7},\; \dfrac{18}{7}\right)$, but constraint 1: $\dfrac{15}{7} + 4 \times \dfrac{18}{7} = \dfrac{87}{7} \approx 12.4 > 8$ ✗

Intersection of constraints 1 and 2 gives $\left(\dfrac{24}{5},\; \dfrac{4}{5}\right)$, but constraint 3: $3 \times \dfrac{24}{5} + \dfrac{4}{5} = \dfrac{76}{5} = 15.2 > 9$ ✗

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Evaluating $z = x + y$ at each corner point:

$A(0,\; 0):\quad z = 0 + 0 = 0$

$B(3,\; 0):\quad z = 3 + 0 = 3$

$C(0,\; 2):\quad z = 0 + 2 = 2$

$D\!\left(\dfrac{28}{11},\; \dfrac{15}{11}\right):\quad z = \dfrac{28}{11} + \dfrac{15}{11} = \dfrac{43}{11}$

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$z_{max} = \dfrac{43}{11} = 3\dfrac{10}{11}$

$\boxed{z_{max} = 3\dfrac{10}{11}}$