The sum of the legs of a right triangle is 9 cm. When the triangle rotates about one of the legs, a cone result which has the maximum volume. Then

Options

A. (a)slant heigth of such a cone is `3sqrt5`
B. (b)maximum value of the cone is `32pi`
C. (c)curved surface of the cone is `18sqrt5pi`
D. (d)semi vertical angle of cone is `tan^(-1)sqrt2`

Answer

To solve the problem, we need to find the dimensions of a right triangle such that when it rotates about one of its legs, the resulting cone has the maximum volume. The sum of the legs of the triangle is given as 9 cm.

### Step-by-Step Solution:

1. **Define the Variables:**
Let the legs of the right triangle be \( r \) and \( h \). According to the problem, we have:
\[ r + h = 9 \]

2. **Express Volume of the Cone:**
When the triangle rotates about one leg, the volume \( V \) of the cone formed is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
We can substitute \( h \) from the first equation into the volume formula. From \( r + h = 9 \), we can express \( h \) as:
\[ h = 9 - r \]
Substituting this into the volume formula gives:
\[ V = \frac{1}{3} \pi r^2 (9 - r) \]

3. **Simplify the Volume Expression:**
Expanding the volume expression:
\[ V = \frac{1}{3} \pi (9r^2 - r^3) \]
To maximize the volume, we can ignore the constant factor \( \frac{1}{3} \pi \) and focus on maximizing:
\[ V' = 9r^2 - r^3 \]

4. **Find the Derivative:**
To find the maximum volume, we take the derivative of \( V' \) with respect to \( r \):
\[ \frac{dV'}{dr} = 18r - 3r^2 \]

5. **Set the Derivative to Zero:**
Setting the derivative equal to zero to find critical points:
\[ 18r - 3r^2 = 0 \]
Factoring out \( 3r \):
\[ 3r(6 - r) = 0 \]
This gives us two solutions:
\[ r = 0 \quad \text{or} \quad r = 6 \]

6. **Determine the Corresponding \( h \):**
If \( r = 6 \):
\[ h = 9 - r = 9 - 6 = 3 \]

7. **Check the Second Derivative:**
To confirm that this critical point gives a maximum, we check the second derivative:
\[ \frac{d^2V'}{dr^2} = 18 - 6r \]
Evaluating at \( r = 6 \):
\[ \frac{d^2V'}{dr^2} = 18 - 36 = -18 < 0 \]
Since the second derivative is negative, \( r = 6 \) gives a local maximum.

8. **Calculate the Volume:**
Now substituting \( r = 6 \) and \( h = 3 \) back into the volume formula:
\[ V = \frac{1}{3} \pi (6^2)(3) = \frac{1}{3} \pi (36)(3) = 36\pi \]

9. **Find the Slant Height:**
The slant height \( l \) of the cone can be found using the Pythagorean theorem:
\[ l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \]

### Conclusion:
The maximum volume of the cone is \( 36\pi \) cm³, and the slant height is \( 3\sqrt{5} \) cm.