William Cleghorn

2021-12-18

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?

What is the ratio now of h to r for the most economical can?

kaluitagf

Beginner2021-12-19Added 38 answers

Step 1

The right circular cylinder's volume is $V=\pi {r}^{2}h$ where h is the height of the cylinder and r is the radius of its circular base. The quantity of aluminum used up by the can must be taken into consideration since waste $A=8{r}^{2}+2\pi rh$. The capacity of the can that can hold the least amount of aluminum is $100c{m}^{3}$, We can only translate A into one variable. Let's translate A into 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 $A=8{r}^{2}+2\pi r\left(\frac{1000}{\pi {r}^{2}}\right)=8{r}^{2}+\frac{2000}{r}$

Let's now solve for r and distinguish A with respect to r by setting it to 0:

$\frac{dA}{dr}=16r-\frac{2000}{{r}^{2}}$

$0=16r-\frac{2000}{{r}^{2}}$

$\frac{2000}{{r}^{2}}=16r$

$16{r}^{3}=2000$

${r}^{3}=125$

$r=\sqrt{3}\left\{125\right\}=5$

Since $h=\frac{1000}{\pi {r}^{2}},\text{}h=\frac{1000}{\pi {\left(5\right)}^{2}}=\frac{40}{\pi}$

Therefore, the ratio of h to r is $\frac{\frac{40}{\pi}}{5}=\frac{8}{\pi}$

(Note: Since $\pi$ is now larger than 2 to 1 and less than 4,)

Natalie Yamamoto

Beginner2021-12-20Added 22 answers

Step 1

$\text{Total area of the can}=\text{area of (top}+\text{bottom)}+\text{lateral area}$

lateral area $2\pi rh$ without waste

If you use a 2r square, the base's area 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 $A\left(r\right)=8{r}^{2}+\frac{2\pi r\left(1000\right)}{\pi {r}^{2}}$

$A\left(r\right)=8{r}^{2}+\frac{2000}{r}$

Considering both derivatives

$A}^{\prime}\left(r\right)=16r-\frac{2000}{{r}^{2}$

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 $h=\frac{1000}{\pi {r}^{2}}$

$h=\frac{1000}{3.14\times 25}$

$h=12.74cm$

ratio $\frac{h}{r}=\frac{12.74}{5}$

$\frac{h}{r}=2.55$

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