To minimize the surface area of a cylinder that is open at one end, we need to consider the two parts of the cylinder that contribute to the surface area: the bottom base and the lateral (side) surface.
First, let's express the cylinder's volume \(\(V\)\) and surface area \(\(A\)\) using the radius \(\(r\)\) and height \(\(h\)\):
The volume \(\(V\)\) of a cylinder is given by the formula:
\(\[V = \pi r^2 h\]\)
Since we're given that \(\(V = 125 \text{ cm}^3\)\), we can express \(\(h\)\) in terms of \(\(r\)\) as:
\(\[h = \frac{V}{\pi r^2}\]\)
\(\[h = \frac{125}{\pi r^2}\]\)
The surface area \(\(A\)\) of a cylinder open at one end consists of the area of the circular base and the area of the side. It can be expressed as:
\(\[A = \pi r^2 + 2\pi r h\]\)
Since \(\(h\)\) is expressed in terms of \(\(r\)\), we can substitute the expression for \(\(h\)\) into the surface area formula:
\(\[A(r) = \pi r^2 + 2\pi r \left(\frac{125}{\pi r^2}\right)\]\)
\(\[A(r) = \pi r^2 + 2\pi r \left(\frac{125}{\pi r^2}\right) = \pi r^2 + \frac{250}{r}\]\)
To minimize the surface area \(\(A(r)\)\), we need to find the value of \(\(r\)\) that minimizes this expression. To do this, we take the derivative \(\(A'(r)\)\) with respect to \(\(r\)\) and set it equal to zero to solve for \(\(r\)\).
\(\[A'(r) = \frac{d}{dr}\left(\pi r^2 + \frac{250}{r}\right)\]\)
\(\[A'(r) = 2\pi r - \frac{250}{r^2}\]\)
Set this derivative equal to zero and solve for \(\(r\)\):
\(\[2\pi r - \frac{250}{r^2} = 0\]\)
\(\[2\pi r^3 = 250\]\)
\(\[r^3 = \frac{250}{2\pi}\]\)
\(\[r^3 = \frac{125}{\pi}\]\)
Now, we solve for \(\(r\)\):
\(\[r = \left(\frac{125}{\pi}\right)^\frac{1}{3}\]\)
After finding the value of \(\(r\)\), we can then plug it back into the formula for \(\(h\)\) to find the corresponding height:
\(\[h = \frac{125}{\pi r^2}\]\)
We can plug the calculated \(\(r\)\) into this formula to find the corresponding \(\(h\)\). Once we have both \(\(r\)\) and \(\(h\)\), we have the dimensions required for the cylinder to have a volume of 125 cm³ with minimal surface area.
Note that we should also check the second derivative to ensure that the value of \(\(r\)\) we've found indeed gives a minimum, not a maximum. The second derivative test would look like this:
\(\[A''(r) = \frac{d^2}{dr^2}\left(\pi r^2 + \frac{250}{r}\right) = 2\pi + \frac{500}{r^3}\]\)
For all \(\(r > 0\)\), which is our domain of interest since a radius can't be negative, \(\(A''(r) > 0\)\), confirming a minimum.
