# You are designing a 1000 cm^{3} right circular cylindrical can w

William Cleghorn 2021-12-18 Answered
You are designing a $1000c{m}^{3}$ right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be
$A=8{r}^{2}+2\pi rh$
What is the ratio now of h to r for the most economical can?
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kaluitagf
Step 1
The volume of a right circular cylinder is $V=\pi {r}^{2}h$ where r is the radius of the circular base of the cylinder and h is the height of the cylinder. Since we have to take waste into account, the amount of aluminum used up by the can $A=8{r}^{2}+2\pi rh$. In order to find the minimum amounts of aluminum for the can hold a volume of $100c{m}^{3}$, we must express A in terms of one variable only. Let's express A in terms of r. So have to express h in terms of r and we know that $\pi {r}^{2}h=1000$. Thus we have:
$\pi {r}^{2}h=1000$
$h=\frac{1000}{\pi {r}^{2}}$
So the surface area in terms of r is $A=8{r}^{2}+2\pi r\left(\frac{1000}{\pi {r}^{2}}\right)=8{r}^{2}+\frac{2000}{r}$
Now let's differentiate A with respect to r and set it to 0 and solve for r:
$\frac{dA}{dr}=16r-\frac{2000}{{r}^{2}}$
$0=16r-\frac{2000}{{r}^{2}}$
$\frac{2000}{{r}^{2}}=16r$
$16{r}^{3}=2000$
${r}^{3}=125$
$r=\sqrt{3}\left\{125\right\}=5$
Since
Therefore, the ratio of h to r is $\frac{\frac{40}{\pi }}{5}=\frac{8}{\pi }$
(Note: Since $\pi$ is less than 4, the new ratio is greater than 2 to 1)

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Natalie Yamamoto
Step 1
$\text{Total area of the can}=\text{area of (top}+\text{bottom)}+\text{lateral area}$
lateral area $2\pi rh$ without waste
area of base (considering that you use 2r square) is $4{r}^{2}$
area of bottom ( for same reason ) $4{r}^{2}$
Then Total area $=8{r}^{2}+2\pi rh$
Now can volume is $1000=\pi {r}^{2}h$
$h=\frac{1000}{\pi {r}^{2}}$
And $A\left(r\right)=8{r}^{2}+\frac{2\pi r\left(1000\right)}{\pi {r}^{2}}$
$A\left(r\right)=8{r}^{2}+\frac{2000}{r}$
Taking derivatives both sides
${A}^{\prime }\left(r\right)=16r-\frac{2000}{{r}^{2}}$
If ${A}^{\prime }\left(r\right)=0$
$16r-\frac{2000}{{r}^{2}}=0$
$\frac{16{r}^{3}-2000}{{r}^{2}}=0$
$16{r}^{3}-2000=0$
${r}^{3}=125$
$r=5cm$ and $h=\frac{1000}{\pi {r}^{2}}$
$h=\frac{1000}{3.14×25}$
$h=12.74cm$
ratio $\frac{h}{r}=\frac{12.74}{5}$
$\frac{h}{r}=2.55$

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