A cylindrical box will be tied up with a ribbon as shown in the accompanying figure. The longest piece of ribbon available is 193 cm long, and 25 cm of that is required for the bow. Find the radius and height of the box with the largest possible volume. The largest box has a radius of \( \square \mathrm{cm} \) and a height of \( \square \mathrm{cm} \).

We are given that the ribbon available is 193 cm long, but 25 cm must be reserved for the bow. Thus the length of ribbon that goes around the box is \[ L = 193-25=168\text{ cm}. \] When the ribbon is wrapped around the cylinder exactly once, it follows a helical (diagonal) path on the lateral surface. Unwrapping this lateral surface gives a rectangle with one side equal to the height \( h \) of the cylinder and the other side equal to the circumference of the base, \( 2\pi r \). The length of the ribbon is the diagonal of this rectangle, so \[ \sqrt{(2\pi r)^2 + h^2}=168. \] Squaring both sides we obtain the constraint \[ (2\pi r)^2 + h^2 = 168^2. \] Our goal is to maximize the volume of the cylinder \[ V=\pi r^2h, \] subject to the constraint \[ h^2+4\pi^2r^2=168^2. \] ### Expressing \( V \) in terms of \( r \) Solve the constraint for \( h \): \[ h^2=168^2-4\pi^2 r^2 \quad\Longrightarrow\quad h=\sqrt{168^2-4\pi^2 r^2}. \] Thus, the volume as a function of \( r \) is \[ V(r)=\pi r^2\sqrt{168^2-4\pi^2r^2}. \] ### Finding the Optimal \( r \) Differentiate \( V(r) \) with respect to \( r \) and set the derivative equal to zero. Let \[ f(r)=\sqrt{168^2-4\pi^2r^2}. \] Then \[ V(r)=\pi r^2 f(r). \] Differentiate using the product rule: \[ V'(r)=\pi \cdot 2r\,f(r)+\pi r^2\,f'(r). \] We now find \( f'(r) \). Since \[ f(r)=\left(168^2-4\pi^2r^2\right)^{\frac{1}{2}}, \] by the chain rule, \[ f'(r)=\frac{1}{2}\left(168^2-4\pi^2r^2\right)^{-\frac{1}{2}}\cdot(-8\pi^2r) =-\frac{4\pi^2r}{\sqrt{168^2-4\pi^2r^2}}. \] Substitute \( f(r) \) and \( f'(r) \) into \( V'(r) \): \[ \begin{aligned} V'(r)&=2\pi r\sqrt{168^2-4\pi^2r^2} + \pi r^2\left(-\frac{4\pi^2r}{\sqrt{168^2-4\pi^2r^2}}\right)\\[1mm] &=2\pi r\sqrt{168^2-4\pi^2r^2} - \frac{4\pi^3r^3}{\sqrt{168^2-4\pi^2r^2}}. \end{aligned} \] Setting \( V'(r)=0 \) (and assuming \( r>0 \)) gives \[ 2\pi r\sqrt{168^2-4\pi^2r^2} = \frac{4\pi^3r^3}{\sqrt{168^2-4\pi^2r^2}}. \] Multiply both sides by \(\sqrt{168^2-4\pi^2r^2}\) to eliminate the denominator: \[ 2\pi r \left(168^2-4\pi^2r^2\right) = 4\pi^3r^3. \] Divide both sides by \(2\pi r\) (since \(r>0\)): \[ 168^2-4\pi^2r^2 = 2\pi^2r^2. \] Solve for \( r^2 \): \[ 168^2 = 6\pi^2r^2 \quad\Longrightarrow\quad r^2=\frac{168^2}{6\pi^2}. \] Notice that \(168^2=28224\). Thus, \[ r^2=\frac{28224}{6\pi^2}=\frac{4704}{\pi^2}, \] and so \[ r=\frac{\sqrt{4704}}{\pi}. \] Factor \(4704\): note that \[ 4704=28^2\cdot6 \quad\Longrightarrow\quad \sqrt{4704}=28\sqrt{6}. \] Thus, the optimal radius is \[ r=\frac{28\sqrt{6}}{\pi}\text{ cm}. \] ### Finding the Corresponding Height \( h \) Return to the constraint \[ h^2=168^2-4\pi^2r^2. \] Substitute \( r^2=\frac{4704}{\pi^2} \): \[ h^2=28224-4\pi^2\cdot\frac{4704}{\pi^2}=28224-4\cdot4704. \] Calculate \[ 4\cdot4704=18816, \] so \[ h^2=28224-18816=9408. \] Taking the square root, \[ h=\sqrt{9408}. \] Factor \(9408\) as follows: \(9408=56^2\cdot3\) (since \(56^2=3136\) and \(3136\cdot3=9408\)) or, more directly, notice that \[ 9408=16\cdot588,\quad 588=49\cdot12,\quad \sqrt{16}=4,\quad \sqrt{49}=7,\quad \sqrt{12}=2\sqrt{3}. \] Thus, \[ h=4\cdot7\cdot2\sqrt{3}=56\sqrt{3}\text{ cm}. \] ### Final Answer The largest box has a radius of \[ \boxed{\frac{28\sqrt{6}}{\pi}\text{ cm}} \] and a height of \[ \boxed{56\sqrt{3}\text{ cm}}. \]