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

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.

We are given $V = 250 \pi \text{ cm}^3$.

$$250 \pi = \pi r^2 h$$

Divide both sides by $\pi$:

$$250 = r^2 h$$

Solve for $h$:

$$h = \frac{250}{r^2}$$

$$A(r) = 2\pi r^2 + 2\pi r \left(\frac{250}{r^2}\right)$$ $$A(r) = 2\pi r^2 + \frac{500\pi r}{r^2}$$ $$A(r) = 2\pi r^2 + \frac{500\pi}{r}$$

To find the minimum area, we need to find the critical points by setting the first derivative to zero.

$$A(r) = 2\pi r^2 + 500\pi r^{-1}$$ $$\frac{dA}{dr} = \frac{d}{dr}(2\pi r^2 + 500\pi r^{-1})$$ $$\frac{dA}{dr} = 2\pi (2r) + 500\pi (-1 r^{-2})$$ $$\frac{dA}{dr} = 4\pi r - \frac{500\pi}{r^2}$$

$$4\pi r - \frac{500\pi}{r^2} = 0$$ $$4\pi r = \frac{500\pi}{r^2}$$

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:

$$r = \sqrt[3]{125}$$ $$r = 5 \text{ cm}$$

$$\frac{d^2A}{dr^2} = \frac{d}{dr}\left(4\pi r - 500\pi r^{-2}\right)$$ $$\frac{d^2A}{dr^2} = 4\pi - 500\pi (-2 r^{-3})$$ $$\frac{d^2A}{dr^2} = 4\pi + \frac{1000\pi}{r^3}$$

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 $\text{5 cm}$.