A cylindrical box will be tied up with a ribbon as shown in the accompanying figure. The longest piece of ribbon available is 202 cm long, and 10 cm of that is required for the bow. Find the radius and height of the box with the largest possible volume.

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To solve this problem, we need to maximize the volume of a cylindrical box given the constraints on the ribbon length.

### Known Conditions:
1. The total length of the ribbon is 202 cm.
2. 10 cm of the ribbon is used for the bow, leaving us with $$ 202 - 10 = 192 $$ cm for the cylindrical box.
3. The ribbon wraps around the cylinder and goes up the height, which means the length of the ribbon can be expressed in terms of the radius $$ r $$ and height $$ h $$ of the cylinder.

### Step 1: Express the ribbon length in terms of $$ r $$ and $$ h $$
The ribbon wraps around the cylinder once, which means the length of the ribbon used can be expressed as:
$$
2\pi r + 2h = 192
$$
This equation accounts for the circumference of the base of the cylinder (which is $$ 2\pi r $$) and the height (which is $$ 2h $$).

### Step 2: Solve for $$ h $$
We can rearrange the equation to express $$ h $$ in terms of $$ r $$:
$$
2h = 192 - 2\pi r
$$
$$
h = 96 - \pi r
$$

### Step 3: Express the volume of the cylinder
The volume $$ V $$ of a cylinder is given by the formula:
$$
V = \pi r^2 h
$$
Substituting $$ h $$ from the previous step:
$$
V = \pi r^2 (96 - \pi r)
$$
$$
V = 96\pi r^2 - \pi^2 r^3
$$

### Step 4: Maximize the volume
To find the maximum volume, we need to take the derivative of $$ V $$ with respect to $$ r $$ and set it to zero:
$$
\frac{dV}{dr} = 192\pi r - 3\pi^2 r^2
$$
Setting the derivative equal to zero:
$$
192\pi r - 3\pi^2 r^2 = 0
$$
Factoring out $$ \pi r $$:
$$
\pi r (192 - 3\pi r) = 0
$$
This gives us two solutions:
1. $$ r = 0 $$ (not valid for a cylinder)
2. $$ 192 - 3\pi r = 0 $$

Solving for $$ r $$:
$$
3\pi r = 192
$$
$$
r = \frac{192}{3\pi} = \frac{64}{\pi}
$$

### Step 5: Find $$ h $$
Now we can substitute $$ r $$ back into the equation for $$ h $$:
$$
h = 96 - \pi \left(\frac{64}{\pi}\right)
$$
$$
h = 96 - 64 = 32
$$

### Conclusion
The dimensions of the cylindrical box that maximize the volume are:
- Radius $$ r = \frac{64}{\pi} $$ cm
- Height $$ h = 32 $$ cm

Now, let's calculate the numerical values for $$ r $$ and $$ h $$.

Calculating $$ r $$:
$$
r \approx \frac{64}{3.14159} \approx 20.37 \text{ cm}
$$

Thus, the final dimensions are:
- Radius $$ r \approx 20.37 $$ cm
- Height $$ h = 32 $$ cm
The cylindrical box with the largest possible volume has a radius of approximately 20.37 cm and a height of 32 cm.

A cylindrical box will be tied up with a ribbon as shown in the accompanying figure. The longest piece of ribbon available is 202 cm long, and 10 cm of that is required for the bow. Find the radius and height of the box with the largest possible volume.

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