
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Let's find the restriction on the area of the field. Let the area of the field be $S$ hectares, then

$$
10 \cdot 300 \cdot S \leq 10000
$$

$$
\mathrm{S} \leq \frac{10}{3}
$$

Also note that the field's area cannot be zero, it consists of one part, not several sections.

Notice that the field is symmetric with respect to the coordinate axes for any parameter value due to the modulus signs.

Consider $|6| y|+3|$, this expression is obviously not less than three.

Then for $2|2| x|-\boldsymbol{a}|$ it is not more than 9, hence $\boldsymbol{a} \geq -4.5$. For smaller values of the parameter, the field does not exist.

Additionally, note that since there is no parameter inside the second modulus, the boundary stretches along the x-axis.

For $a > 4.5$, the point $(0; 0)$ does not lie within the bounded region, making the region consist of two parts, contradicting the logical sense of the problem.

Thus, we have the range for $a$ from $-4.5$ to $4.5$, considering only integer values, we have the finite set of parameters: $\pm 4, \pm 3, \pm 2, \pm 1, 0$.

Proceeding in increasing order, let $a = -4$:

$$2(2 x + 4) + (6 y + 3) = 12$$

$$\mathrm{x} = 0, \mathrm{y} = \frac{1}{6}$$

$$\mathrm{y} = 0, \mathrm{x} = \frac{1}{4}$$

$$\mathrm{s} = 2 \cdot \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{12} < \frac{10}{3}$$

Let $a = -3$:

$$4 x + 6 + 6 y + 3 = 12$$

$$4 x + 6 y = 3$$

$$\mathrm{x} = 0, \mathrm{y} = \frac{1}{2}$$

$$\mathrm{y} = 0, \mathrm{x} = \frac{3}{4}$$

$$\mathrm{s} = 2 \cdot \frac{1}{2} \cdot \frac{3}{4} = \frac{3}{4} < \frac{10}{3}$$

Let $a = -2$:

$$4 x + 4 + 6 y + 3 = 12$$

$$4 x + 6 y = 5$$

$$x = 0, y = \frac{5}{6}$$

$$y = 0, x = \frac{5}{4}$$

$$\mathrm{s} = 2 \cdot \frac{5}{6} \cdot \frac{5}{4} = \frac{25}{12} < \frac{10}{3}$$

Let $a = -1$:

$$4 x + 2 + 6 y + 3 = 12$$

$$4 x + 6 y = 7$$

$$x = 0, y = \frac{7}{6}$$

$$y = 0, x = \frac{7}{4}$$

$$\mathrm{s} = 2 \cdot \frac{7}{6} \cdot \frac{7}{4} = \frac{49}{12} > \frac{10}{3}$$

So, it does not fit the criteria.

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