Let's use related rates to solve this problem.
We are given that the radius of the conical water tank is 8 meters and the height is 12 meters. We want to find how fast the water level is rising, which means we need to find the rate of change of the height, dh/dt, when the water level is 6 meters deep.

First, we need to establish a relationship between the height and the radius of the cone at any given time.

The volume of a cone is given by the formula V = (1/3) * π * r^2 * h, where V is the volume, r is the radius, and h is the height.

We can differentiate this equation implicitly with respect to time t:

dV/dt = (1/3) * π * (2r * dr/dt) * h + (1/3) * π * r^2 * dh/dt.

We are given dV/dt = 4 cubic meters per minute, and we are asked to find dh/dt when h = 6 meters.

Let's solve for dr/dt using the given radius:

4 = (1/3) * π * (2 * 8 * dr/dt) * 6 + (1/3) * π * 8^2 * dh/dt.
4 = (16/3) * π * dr/dt * 6 + (64/3) * π * dh/dt.

Now, let's substitute h = 6 into the equation:

4 = (16/3) * π * dr/dt * 6 + (64/3) * π * dh/dt.
4 = 32 * π * dr/dt + (64/3) * π * dh/dt.

Since the water level is halfway up the tank when h = 6 meters, the radius of the water surface, r, can be found using similar triangles. The ratio of the radius of the water surface to the radius of the tank is equal to the ratio of the height of the water level to the height of the tank.

Using the given radius and height, we can calculate the radius of the water surface when h = 6:

r = (8/12) * 6 = 4 meters.

Substituting r = 4 into the equation:

4 = 32 * π * dr/dt + (64/3) * π * dh/dt.

We can cancel out the π term:

4 = 32 * 4 * dr/dt + (64/3) * dh/dt.
4 = 128 * dr/dt + (64/3) * dh/dt.

Simplifying the equation:

128 * dr/dt = 4 - (64/3) * dh/dt.
dr/dt = (4 - (64/3) * dh/dt) / 128.

Now, we know that when h = 6, dh/dt = ?.

Let's plug in h = 6 and solve for dr/dt:

dr/dt = (4 - (64/3) * dh/dt) / 128,
dr/dt = (4 - (64/3) * dh/dt) / 128,
dr/dt = (4 - (64/3) * (dh/dt)) / 128,
dr/dt = (4 - (64/3) * dh/dt) / 128.

We can substitute 4/128 for 1/32 and simplify:

dr/dt = (1/32 - (2/3) * (dh/dt)).

To find dh/dt, we need to substitute dr/dt and h = 6 into the following equation:

4 = 32 * π * dr/dt + (64/3) * π * dh/dt.

Substituting dr/dt = (1/32 - (2/3) * (dh/dt)):

4 = 32 * π * ((1/32 - (2/3) * (dh/dt))) + (64/3) * π * dh/dt,
4 = 1 - (2/3) * (32 * π * dh/dt) + (64/3) * π * dh/dt,
4 = 1 - (64/3) * π * (dh/dt) + (64/3) * π * dh/dt,
4 = (64/3) * π * dh/dt - (64/3) * π * (dh/dt) + 1,
4 = (64/3) * π * dh/dt - (64/3) * π * dh/dt + 1,
4 = 1.

This equation is not true, so there must have been an error in our calculations.
Let's try a different approach.

When the tank is half full, the radius of the water surface is half the radius of the tank. Since the radius of the tank is 8 meters, the radius of the water surface is 4 meters.

We can use similar triangles to find the height of the water level, h, when the radius of the water surface is 4 meters:

h/12 = 4/8,
h = 6 meters.

Now, we can find dh/dt when h = 6 meters:

V = (1/3) * π * r^2 * h,
4 = (1/3) * π * 8^2 * 6,
4 = (128/3) * π.

We can differentiate this equation implicitly with respect to time t:

dV/dt = (128/3) * π * (dh/dt).

We are given dV/dt = 4 cubic meters per minute, so:

4 = (128/3) * π * (dh/dt).

Now, we can solve for dh/dt:

dh/dt = 4 / ((128/3) * π).

Simplifying the expression:

dh/dt = 4 / (128/3 * π),
dh/dt = 12/128 * 3/π,
dh/dt ≈ 0.088 meters per minute.

Therefore, the water level is rising at a rate of approximately 0.088 meters per minute when the water is 6 meters deep.

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