To solve this problem, we aim to find the altitude of the inscribed cone that maximizes its volume. Let's break it down step by step:

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1. Outer cone specifications:

• Volume: $V_{\text{outer}} = 6 \, \text{m}^3$
• Altitude: $h_{\text{outer}} = 7 \, \text{m}$

From the volume formula for a cone, $V = \frac{1}{3} \pi r^2 h$, the radius of the base of the outer cone is:

$V_{\text{outer}} = \frac{1}{3} \pi r_{\text{outer}}^2 h_{\text{outer}}$ $6 = \frac{1}{3} \pi r_{\text{outer}}^2 (7)$ $r_{\text{outer}}^2 = \frac{18}{7 \pi} \implies r_{\text{outer}} = \sqrt{\frac{18}{7 \pi}}$

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2. Geometry of the inscribed cone:

The inscribed cone shares the same axis with the outer cone, and its vertex is on the base of the outer cone. If the altitude of the inscribed cone is $h$, the radius of its base is proportional to its altitude because the cones are similar in shape. Let the ratio of similarity be $k$:

$k = \frac{r_{\text{inscribed}}}{r_{\text{outer}}} = \frac{h}{h_{\text{outer}}} = \frac{h}{7}$

Thus, the radius of the inscribed cone is: $r_{\text{inscribed}} = k \cdot r_{\text{outer}} = \frac{h}{7} \cdot r_{\text{outer}}$

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3. Volume of the inscribed cone:

The volume $V_{\text{inscribed}}$ of the inscribed cone is: $V_{\text{inscribed}} = \frac{1}{3} \pi r_{\text{inscribed}}^2 h$ Substitute $r_{\text{inscribed}} = \frac{h}{7} \cdot r_{\text{outer}}$: $V_{\text{inscribed}} = \frac{1}{3} \pi \left( \frac{h}{7} \cdot r_{\text{outer}} \right)^2 h$ $V_{\text{inscribed}} = \frac{1}{3} \pi \left( \frac{h}{7} \right)^2 r_{\text{outer}}^2 h$ $V_{\text{inscribed}} = \frac{1}{3} \pi \frac{h^3}{49} r_{\text{outer}}^2$

Substitute $r_{\text{outer}}^2 = \frac{18}{7 \pi}$: $V_{\text{inscribed}} = \frac{1}{3} \pi \frac{h^3}{49} \cdot \frac{18}{7 \pi}$ $V_{\text{inscribed}} = \frac{6}{343} h^3$

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4. Maximizing $V_{\text{inscribed}}$:

To maximize $V_{\text{inscribed}}$, differentiate it with respect to $h$ and set the derivative to zero: $V_{\text{inscribed}} = \frac{6}{343} h^3$ $\frac{dV_{\text{inscribed}}}{dh} = \frac{18}{343} h^2$ Set $\frac{dV_{\text{inscribed}}}{dh} = 0$: $\frac{18}{343} h^2 = 0 \implies h = 0 \text{ or } h = 7$

For the volume to be maximized within the cone, $h$ cannot be $0$. The maximum occurs at $h = 7 \, \text{m}$, but $h = 7$ corresponds to the outer cone. To ensure the cone is inscribed, let's confirm $h < 7$.

5. Confirming Geometry and Limits

Actually, I realized a small misstep. Let's revisit and properly conclude the condition.

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Since $h$ must be less than 7 (as the inscribed cone cannot extend beyond the outer cone), the critical step is in correctly setting up our optimization problem using geometry and derivatives.

For simplicity:

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