A close cylindrical metal tin is to have a capacity of 250 pi ml. If the area of the metal used is to be a minimum comma what should the radius of the tin be?

To find the radius of the tin that will minimize the surface area of the metal used, we need to set up an equation for the surface area of the cylinder in terms of the radius.

The surface area of a closed cylindrical tin is given by:

A = 2πr^2 + 2πrh

Where r is the radius of the tin and h is the height of the tin.

Given that the capacity of the tin is 250π ml, we can write:

V = πr^2h = 250π

Solving for h, we get:

h = 250/r^2

Substitute this expression for h into the equation for the surface area:

A = 2πr^2 + 2πr(250/r^2)

A = 2πr^2 + 500π/r

To find the radius that minimizes the surface area, we need to find the critical points of A by taking the derivative with respect to r and setting it equal to zero:

dA/dr = 4πr - 500π/r^2 = 0

4πr = 500π/r^2

r^3 = 125

r = 5

Therefore, the radius of the tin that will minimize the surface area of the metal used is 5 units.

Let the radius and height of the cylinder be r and h, respectively. Then

Volume of the cylinder = 250 pi ml = 250 ml
$$pi r^2 h = 250 ml$$
Surface area of the cylinder = 2 pi r(r + h) cm^2

We want to minimize the surface area, so we can write:

$$S = 2 pi r(r + h)$$
$$= 2 pi r(r + 250/(pi r^2))$$
$$= 2 pi r(r + 800/(pi r^3))$$
$$\frac{dS}{dr} = 2 pi (r + 800/(pi r^3)) + 2 pi r (-3\times 800/(pi r^4))$$
$$\frac{dS}{dr} = 2 pi (r - 2400/(pi r^4))$$

For minima or maxima, $$\frac{dS}{dr} = 0$$
$$2 pi (r - 2400/(pi r^4)) = 0$$
$$r - 2400/(pi r^4) = 0$$
$$r^5 = 2400/pi$$
$$r^5 = 2400\times 7/22$$
$$r = (2400\times 7/22)^{1/5}$$
$$r = 60^{1/5}$$
$$r = 3.22 cm$$