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