The maximum value of the objective function $z = 10x + 15y$ of an L.P.P. subjected to the constraints

$2x + 4y \leq 8$,

$3x + y \leq 6$,

$-x - y \geq -4$,

$x \geq 0$, $y \geq 0$ is:

The constraints can be simplified as:

$2x + 4y \leq 8 \implies x + 2y \leq 4$

$3x + y \leq 6$

$-x - y \geq -4 \implies x + y \leq 4$

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

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Corner point $A$: $x = 0$ and $y = 0$

$A = (0,\; 0)$

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Corner point $B$: $y = 0$ and $3x + y = 6$

$3x = 6$

$x = 2$

$B = (2,\; 0)$

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Corner point $C$: $x + 2y = 4$ and $3x + y = 6$

From the first equation: $x = 4 - 2y$

$3(4 - 2y) + y = 6$

$12 - 6y + y = 6$

$-5y = -6$

$y = \dfrac{6}{5}$

$x = 4 - 2!\left(\dfrac{6}{5}\right)

= 4 - \dfrac{12}{5}

= \dfrac{8}{5}$

$C = \left(\dfrac{8}{5},\; \dfrac{6}{5}\right)$

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Corner point $D$: $x = 0$ and $x + 2y = 4$

$2y = 4$

$y = 2$

$D = (0,\; 2)$

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Evaluating $z = 10x + 15y$ at each corner point:

At $A(0,\; 0)$: $z = 10(0) + 15(0) = 0$

At $B(2,\; 0)$: $z = 10(2) + 15(0) = 20$

At $C\!\left(\dfrac{8}{5},\; \dfrac{6}{5}\right)$: $z = 10\!\left(\dfrac{8}{5}\right) + 15\!\left(\dfrac{6}{5}\right) = 16 + 18 = 34$

At $D(0,\; 2)$: $z = 10(0) + 15(2) = 30$

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The maximum value of $z$ is $\boxed{34}$, occurring at $\left(\dfrac{8}{5},\; \dfrac{6}{5}\right)$.