Question

Of all the right circular cylindrical cans of volume $128\pi c{{m}^{3}}$ find the dimensions of the can which has minimal surface area

Answer 1

Now we are given that right circular cylindrical cans have volume $128\pi c{{m}^{3}}$. We know that the volume of the cylinder is given by $\pi {{r}^{2}}h$. Hence using this we will get a relation in r and h.
Now the curved surface area of the cylinder is $2\pi rh$ and the area of circular disc is $\pi {{r}^{2}}$ . Hence the total surface area of the cylinder will become $2\pi {{r}^{2}}+2\pi rh$. Now since we know the relation between r and h we will convert the formula in one variable.
Now we have the formula for Surface area S. we know the condition to make it maximum or minimum is $\dfrac{ds}{dr}=0$. Hence using this we will find r for which the condition is extrema. And then with the second derivative text we can find if it's maximum or minimum.

Complete step-by-step answer:
Now consider the cylinder whose radius is r and height is h

Now the volume of cylinder is $128\pi c{{m}^{3}}$ and the formula for Volume of cylinder is $V=\pi {{r}^{2}}h$
Hence we get

& 128\pi =\pi {{r}^{2}}h \\\ & \Rightarrow 128={{r}^{2}}h \\\ & \Rightarrow h=\dfrac{128}{{{r}^{2}}}................(1) \\\ \end{aligned}$$ Now we know that the curved surface area of the cylinder is $2\pi rh$ and the area of circular disc is $\pi {{r}^{2}}$ . Hence the total surface area of the cylinder will become $2\pi {{r}^{2}}+2\pi rh$ . Let S be the total surface area of cylinder $S=2\pi {{r}^{2}}+2\pi rh$ $S=2\pi {{r}^{2}}+2\pi r\left( \dfrac{128}{{{r}^{2}}} \right)$ $S=2\pi {{r}^{2}}+\dfrac{256\pi }{r}$ Now let us differentiate above equation and equate to 0 to get the condition for extrema. $\dfrac{dS}{dr}=2\pi (2r)+256\pi \left( -\dfrac{1}{{{r}^{2}}} \right)$ Now equating this to 0 we get $\dfrac{dS}{dr}=0\Rightarrow 2\pi (2r)=256\pi \left( -\dfrac{1}{{{r}^{2}}} \right)$ $\begin{aligned} & \Rightarrow 4\pi r=\dfrac{256\pi }{{{r}^{2}}} \\\ & \Rightarrow 4{{r}^{3}}=256 \\\ & \Rightarrow {{r}^{3}}=64 \\\ & \Rightarrow r=\sqrt[3]{64}=4 \\\ \end{aligned}$ Hence we will get extrema at r = 4. And if r = 4 we have $h=\dfrac{128}{{{r}^{2}}}$ $h=\dfrac{128}{{{4}^{2}}}=\dfrac{128}{16}=8$ Now again consider $\dfrac{dS}{dr}=2\pi (2r)+256\pi \left( -\dfrac{1}{{{r}^{2}}} \right)$ Now let us check the second derivative $\dfrac{{{d}^{2}}s}{d{{r}^{2}}}=2\pi (2)+256\pi \left( \dfrac{2}{{{r}^{3}}} \right)$ Now at $r=4,h=8$ we get $\begin{aligned} & \dfrac{{{d}^{2}}s}{d{{r}^{2}}}=4\pi +256\pi \left( \dfrac{2}{({{4}^{3}})} \right) \\\ & =4\pi +256\pi \left( \dfrac{2}{64} \right) \\\ & =4\pi +4\pi (2) \\\ & =4\pi +8\pi \\\ & =12\pi \\\ \end{aligned}$ Since $\dfrac{{{d}^{2}}s}{d{{r}^{2}}}>0$ we will have a minima at this point Hence the required dimension to get minimum surface area is $r=4,h=8$ **Note:** Now the curved surface area of the cylinder is $2\pi rh$ and this is not the total surface area as we need to consider the top and bottom circular area for total surface area. Hence we get the total surface area is $\pi {{r}^{2}}+2\pi rh$ .