We are tasked with finding the new radius $r$ for the cone after increasing its height by $26\%$, while keeping the total surface area $S$ constant.

Step-by-step approach:

Given:

• Height of the cone: $h = 8 \, \text{cm}$
• Radius of the base: $r = 6 \, \text{cm}$
• The formula for total surface area $S$ of the cone: $S = \pi r^2 + \pi r \sqrt{r^2 + h^2}$
• The height increases by $26\%$, so the new height is: $h_{\text{new}} = h \times (1 + 0.26) = 8 \times 1.26 = 10.08 \, \text{cm}$

Let the new radius be $r_{\text{new}}$.

Step 1: Define the function for surface area

The surface area is a function of $r$ and $h$: $S(r, h) = \pi r^2 + \pi r \sqrt{r^2 + h^2}$

Step 2: Find the derivative with respect to $r$

To estimate the new radius using the tangent line approximation, we need to find the derivative of $S$ with respect to $r$. This derivative is given by: $\frac{dS}{dr} = \frac{d}{dr} \left( \pi r^2 + \pi r \sqrt{r^2 + h^2} \right)$

Using the product rule and chain rule, the derivative becomes: $\frac{dS}{dr} = 2\pi r + \pi \sqrt{r^2 + h^2} + \pi r \cdot \frac{r}{\sqrt{r^2 + h^2}}$

Simplify the derivative: $\frac{dS}{dr} = 2\pi r + \pi \left( \sqrt{r^2 + h^2} + \frac{r^2}{\sqrt{r^2 + h^2}} \right)$

Step 3: Evaluate the derivative at the initial point

Substitute $r = 6 \, \text{cm}$ and $h = 8 \, \text{cm}$ into the derivative to find the slope at this point:

• $r = 6$
• $h = 8$
• $\sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Now calculate the derivative at $r = 6$: $\frac{dS}{dr} = 2\pi(6) + \pi \left( 10 + \frac{6^2}{10} \right)$ $= 12\pi + \pi \left( 10 + \frac{36}{10} \right)$ $= 12\pi + \pi \left( 10 + 3.6 \right)$ $= 12\pi + \pi \times 13.6$ $= 12\pi + 13.6\pi = 25.6\pi$

Thus, the derivative at $r = 6$ is $25.6\pi$.

Step 4: Estimate the new radius using the tangent line approximation

The tangent line approximation for the change in surface area is: $\Delta S \approx \frac{dS}{dr} \times \Delta r$ Where $\Delta r$ is the change in radius.

Since the surface area is constant, $\Delta S = 0$. So, the change in radius $\Delta r$ is given by: $0 = \frac{dS}{dr} \times \Delta r$ Thus, the new radius $r_{\text{new}}$ can be approximated by: $r_{\text{new}} \approx r + \frac{1}{\frac{dS}{dr}} \times \Delta h$

Using $r = 6 \, \text{cm}$, $h = 8 \, \text{cm}$, $h_{\text{new}} = 10.08$ cm and $r$new, thus