Nilai maksimum fungsi objektif Z = 12x + 8y pada daerah penyelesaian dari sistem pertidaksamaan yang ditunjukkan grafik tersebut adalah ....

Line 1 passes through $(20, 0)$ and $(0, 16)$. The equation is $\frac{x}{20} + \frac{y}{16} = 1$.
Multiplying by 80 (LCM of 20 and 16), we get $4x + 5y = 80$.

Line 2 passes through $(10, 0)$ and $(0, -8)$. The equation is $\frac{x}{10} + \frac{y}{-8} = 1$.
Multiplying by 40 (LCM of 10 and 8), we get $4x - 5y = 40$.

The vertices of the shaded triangular region are:

1. The y-intercept of Line 1: $(0, 16)$
2. The x-intercept of Line 2: $(10, 0)$
3. The intersection point of Line 1 and Line 2.

We have the system of equations:

Add equation (1) and equation (2):

Substitute $x=15$ into equation (1):

So, the third vertex is $(15, 4)$.

1. At $(0, 16)$: $$Z = 12(0) + 8(16) = 0 + 128 = 128$$
2. At $(10, 0)$: $$Z = 12(10) + 8(0) = 120 + 0 = 120$$
3. At $(15, 4)$: $$Z = 12(15) + 8(4) = 180 + 32 = 212$$

Comparing the values 128, 120, and 212, the maximum value is 212.

The final answer is $\boxed{\text{D. } 212}$.

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