To maximize the lateral surface area of an inverted cone with a volume of 27, use the following formulas: Volume (V): \(V = \frac{1}{3}\pi r^2 h\) Lateral Surface Area (SA): \(SA = \pi r \sqrt{r^2 + h^2}\) You have already found that \(h = \frac{81}{\pi r^2}\) by solving the volume formula for \(h\): \(27 = \frac{1}{3}\pi r^2 h\) Now, to proceed, substitute \(h\) into the lateral surface area formula: \(SA = \pi r \sqrt{r^2 + \left(\frac{81}{\pi r^2}\right)^2}\) \(SA = \pi r \sqrt{r^2 + \frac{6561}{\pi^2 r^4}}\) To find the maximum of the surface area, take the derivative of SA with respect to \(r\) and set it to zero to solve for \(r\). Then, use this value of \(r\) to find the corresponding height \(h\), and finally substitute both \(r\) and \(h\) back into the lateral surface area formula to find the maximum value.

To maximize the lateral surface area (SA) of the inverted cone, we need to first take the derivative of SA with respect to \(r\), set it to zero and solve for \(r\). The lateral surface area as a function of \(r\) is given by:

\[SA(r) = \pi r \sqrt{r^2 + \frac{6561}{\pi^2 r^4}}\]

Differentiate SA with respect to \(r\):

\[\frac{d(SA)}{dr} = \pi \sqrt{r^2 + \frac{6561}{\pi^2 r^4}} + \pi r \frac{\frac{d}{dr}(r^2 + \frac{6561}{\pi^2 r^4})}{2\sqrt{r^2 + \frac{6561}{\pi^2 r^4}}}\]

Now we need to differentiate the term inside the square root:

\[\frac{d}{dr}(r^2 + \frac{6561}{\pi^2 r^4}) = 2r - \frac{6561 \times 4}{\pi^2 r^5}\]

So, the derivative becomes:

\[\frac{d(SA)}{dr} = \pi \sqrt{r^2 + \frac{6561}{\pi^2 r^4}} + \pi r \frac{2r - \frac{26244}{\pi^2 r^5}}{2\sqrt{r^2 + \frac{6561}{\pi^2 r^4}}}\]

The maximum occurs when the derivative is equal to zero:

\[0 = \pi \sqrt{r^2 + \frac{6561}{\pi^2 r^4}} + \pi r \frac{2r - \frac{26244}{\pi^2 r^5}}{2\sqrt{r^2 + \frac{6561}{\pi^2 r^4}}}\]

This equation is complicated and solving it analytically for \(r\) isn't straightforward. Typically, one would use numerical methods or graphing calculators to find the approximate value of \(r\) that satisfies this condition. Once \(r\) is found, we can substitute it back into the expression for \(h\):

\[h = \frac{81}{\pi r^2}\]

to find the corresponding height of the cone. Then, with the values of \(r\) and \(h\), we can finally substitute both back into the lateral surface area formula to find the maximum lateral surface area.

Extra: The lateral surface area (SA) of an inverted cone describes the area of the cone's side, not including the base. The volume of a cone relates the amount of three-dimensional space it occupies, and in this case, it is given to be 27 cubic units.

Finding the maximum or minimum of a function using calculus involves taking the derivative, which gives the rate of change of the function, and setting it to zero. This is because at the maximum or minimum point, the slope of the tangent to the function is zero (it is either perfectly horizontal if the function is smooth at that point or there is a cusp or corner).

After finding the critical points where the derivative is zero, one would typically use the second derivative test to determine whether these points are maxima, minima, or points of inflection. However, the process described above is specifically focused on maximizing the lateral surface area, assuming that the found critical points correspond to a maximum because of the physical context (a closed shape like a cone typically has a well-defined maximum surface area for given constraints).

The volume formula for a cone \((V = \frac{1}{3}\pi r^2 h)\) tells us that the volume is directly proportional to both the square of the radius of the base and the height of the cone. Maximizing the lateral surface area while maintaining a constant volume is a typical problem in optimization, where certain parameters must be adjusted within a set of constraints to find the best possible outcome - in this case, the largest lateral surface area.