Maximizing The Cylinder's Surface Area

We know that the volume of a cylinder is $\pi r^2 h$ so $\pi r^2 h =900$ . We can then rearrange for h, $h= \frac{900}{\pi r^2}$ . We must now find a formula for the total surface area (S) of a cylinder which is just the bottom and top areas plus the curved surface area so $S=2\pi r^2 +2\pi r h$ . We can then use our equation for h to rewrite the equation for the surface area so that it only depends upon the radius, $S=2\pi r^2+\frac{2\cdot 900 \cdot \pi \cdot r}{\pi \cdot r^2}=2\pi r^2+\frac{1800}{r}$ . We can differentiate this with respect to r once and set equal to 0 to find any stationary points (minimums,maximums). $\frac{dS}{dr}=4\pi r -\frac{1800}{r^2}=0 \implies \frac{1800}{r^2}=4\pi r \implies 450= \pi r^3 \implies r= \sqrt[3]{\frac{450}{\pi}}$ . There is therefore only a single stationary point. If we take the second derivative of S with respect to r we can determine this stationary points nature i.e. whether it is a maximum or minimum. $\frac{d^2S}{dr^2}=4 \pi+\frac{3600}{r^3}$ . If we now plug in the value of r at our stationary point we get $\frac{d^2S}{dr^2}_{r=\frac{450}{\pi}}=4 \pi + \frac{3600}{\frac{450}{\pi}}=4 \pi +8 \pi =12 \pi$ . $\frac{d^2S}{dr^2}>0 \implies$ that the stationary point is a minimum as we have only one stationary point the surface area has no maximum i.e. The maximum surface area does not exist.

We have that $V= \pi r^2h=900$ . We know that the surface area of the cylinder is $S=2 \pi rh + 2 \pi r^2$ . From the equation for Volume we know that $\pi rh=\frac{900}{ r}$ and $\pi r^2=\frac{900}{h}$ . Making these substitutions in to the Surface Area equation yields:

$S=1800 \left(\frac{1}{r}+\frac{1}{h} \right)$

In order to maximise the surface area, we must maximise the value of $\frac{1}{r}+\frac{1}{h}$ . This is done by minimising the values of $r$ and $h$ . We note that as $r \rightarrow 0$ and $h \rightarrow 0$ , $\frac{1}{r}+\frac{1}{h} \rightarrow \infty$ . This means that the radius and height can be made infinitely small and the surface area can be made infinitely large implying that there is no maximum surface area. Hence the answer is: $\boxed{No, \ because \ the \ maximum \ surface \ area \ does \ not \ exist.}$

Basically, you can make a cylinder with a radius of infinite size and a height that is arbitrarily small. Thus, since the radius is infinitely long, our two bases have an infinite surface area, which proves that our cylinder has no maximum surface area.

The volume of a cylinder is given by $V=\pi r^2h$ . The surface area is given as the sum of the top, bottom and side area, $A=2\pi r^2+2\pi rh$ . Since $V$ is fixed ( $V=900 \, cm^3$ ), we have $h=\frac{V}{\pi r^2}$ . Substitution into the formula for A gives $A=\frac{2V}{r}+2 \pi r^2$ Now we observe that $\lim_{r \to +\infty} A=\infty$ (which is also true as $r \to 0^+$ , but that doesn't matter). This means that the surface area gets arbitrarily big as $r \to +\infty$ , so there is no maximum area (the solution for $\frac{\mathrm{d} A}{\mathrm{d} r}=0$ is a minimum, which can be verified).

Total Surface area = 2 pi r h + 2 pi r^2. and volume = pi * r^2 * h =900. After substituting h in terms of r and On double differentiating, we get : 3600/(r^3). Now if double differentiation gives positive value, then no maxima can be possible.

First, the surface area could be expressed as $2\pi r^2 + 2\pi rh$

As $V = \pi r^2h = 900$ , $h = \frac{900}{r}$

The formula can then be re-written as $2\pi r^2h + \frac{1800}{r}$

Differentiating this gives $4\pi r - \frac{1800}{r^2}$ , the second derivative gives $4\pi + \frac{3600}{r^3}$

As this is always positive because $r$ must be positive, this outputs a minimum surface area and we must find a way of optimizing.

The maximum of $r$ is $\sqrt{\frac{450}{\pi}}$ where the surface area consists of only the 2 circles and where the height is zero.

The maximum of $2\pi r^2h + \frac{1800}{r}$ in the interval $0 < r < \sqrt{\frac{450}{\pi}}$ is infinity as $r$ approaches 0.

Therefore, there is no maximum surface area.

Maximizing the surface area requires a change in the volume of the cylinder, which is fixed in this case.

Increasing the surface area of a cylinder will either increase its volume or deform the cylinder itself. Hence, for a given volume of a cylinder, its surface area is unique, & no 'maximum surface area" exists.

As r gets very big, S approaches infinity. Similarly, as r gets very small, S also approaches infinity.

To visualize this, if you squash the cylinder so that it has a tiny height, the bases can be expanded to keep the volume constant. Since the bases can be expanded to any size, the maximum surface area is infinite. If you squeeze the cylinder so that it has a tiny radius but large height, surface area would also increase, but it's a bit harder to see this intuitively