To find the radius that minimizes the metal used for a closed cylindrical tin with a given capacity, we need to use calculus.

To find the radius that minimizes the metal used for a closed cylindrical tin with a given capacity, we need to use calculus.

Given:

• Capacity (Volume) of the cylinder, $V = 250 \pi \text{ ml}$. We assume $1 ml = 1 cm^3$, so $V = 250 \pi cm^3$.
• Let $r$ be the radius and $h$ be the height of the cylinder.

Step 1: Write down the formulas for the volume and surface area of a closed cylinder. The volume of a cylinder is given by: $V = \pi r^2 h$ The surface area of a closed cylinder (area of metal used) is given by: $A = 2\pi r^2 + 2\pi rh$ The $2\pi r^2$ accounts for the top and bottom circular bases, and $2\pi rh$ accounts for the curved side.

Step 2: Express the height ($h$) in terms of the radius ($r$) using the given volume. We are given $V = 250 \pi cm^3$. $250 \pi = \pi r^2 h$ Divide both sides by $\pi$: $250 = r^2 h$ Solve for $h$: $h = \frac{250}{r^2}$

Step 3: Substitute the expression for $h$ into the surface area formula to get $A$ as a function of $r$ only. $A(r) = 2\pi r^2 + 2\pi r \left(\frac{250}{r^2}\right)$ $A(r) = 2\pi r^2 + \frac{500\pi r}{r^2}$ $A(r) = 2\pi r^2 + \frac{500\pi}{r}$

Step 4: Differentiate the area function $A(r)$ with respect to $r$. To find the minimum area, we need to find the critical points by setting the first derivative to zero. $A(r) = 2\pi r^2 + 500\pi r^{-1}$ $\frac{dA}{dr} = \frac{d}{dr}(2\pi r^2 + 500\pi r^{-1})$ $\frac{dA}{dr} = 2\pi (2r) + 500\pi (-1 r^{-2})$ $\frac{dA}{dr} = 4\pi r - \frac{500\pi}{r^2}$

Step 5: Set the derivative $\frac{dA}{dr}$ to zero and solve for $r$. $4\pi r - \frac{500\pi}{r^2} = 0$ $4\pi r = \frac{500\pi}{r^2}$ Divide both sides by $\pi$: $4r = \frac{500}{r^2}$ Multiply both sides by $r^2$: $4r^3 = 500$ $r^3 = \frac{500}{4}$ $r^3 = 125$ Take the cube root of both sides: $r = \sqrt[3]{125}$ $r = 5 \text{ cm}$

Step 6: (Optional) Verify that this radius corresponds to a minimum area using the second derivative test. $\frac{d^2A}{dr^2} = \frac{d}{dr}\left(4\pi r - 500\pi r^{-2}\right)$ $\frac{d^2A}{dr^2} = 4\pi - 500\pi (-2 r^{-3})$ $\frac{d^2A}{dr^2} = 4\pi + \frac{1000\pi}{r^3}$ Substitute $r=5$: $\frac{d^2A}{dr^2} = 4\pi + \frac{1000\pi}{(5)^3}$ $\frac{d^2A}{dr^2} = 4\pi + \frac{1000\pi}{125}$ $\frac{d^2A}{dr^2} = 4\pi + 8\pi$ $\frac{d^2A}{dr^2} = 12\pi$ Since $12\pi > 0$, the surface area is indeed a minimum at $r=5 \text{ cm}$.

The radius of the tin that minimizes the metal used is $\boxed{5 cm}$.