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

Step 1
The right circular cylinder's volume is ${\displaystyle V=\pi r^{2}h}$ 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 ${\displaystyle A=8r^2+2\pi rh}$. The capacity of the can that can hold the least amount of aluminum is ${\displaystyle 100c{m}^3}$, 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 ${\displaystyle \pi r^{2}h=1000}$. Thus we have:
${\displaystyle \pi r^{2}h=1000}$
${\displaystyle h=\frac{1000}{\pi r^2}}$
So the surface area in terms of r is ${\displaystyle A=8r^2+2\pi r\left(\frac{1000}{\pi r^2}\right)=8r^2+\frac{2000}{r}}$
Let's now solve for r and distinguish A with respect to r by setting it to 0:
${\displaystyle \frac{dA}{dr}=16r-\frac{2000}{r^2}}$
${\displaystyle 0=16r-\frac{2000}{r^2}}$
${\displaystyle \frac{2000}{r^2}=16r}$
${\displaystyle 16r^3=2000}$
${\displaystyle r^3=125}$
${\displaystyle r=\sqrt{3}\left\{125\right\}=5}$
Since ${\displaystyle h=\frac{1000}{\pi r^2},h=\frac{1000}{\pi {\left(5\right)}^2}=\frac{40}{\pi }}$
Therefore, the ratio of h to r is ${\displaystyle \frac{\frac{40}{\pi }}{5}=\frac{8}{\pi }}$
(Note: Since ${\displaystyle \pi }$ is now larger than 2 to 1 and less than 4,)

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