An industrial mixer is made up of an open cylinder connected to a cone. The height of the cylinder and the cone is 1.2 m and 0.5 m respectively. The diameter of the mixer is 1.5 m.

1. Find the radius of the mixer from its diameter.

Diameter $=1.5$ m

Radius $=\dfrac{1.5}{2}=0.75$ m

2. Calculate the volume of the cylindrical part.

Volume of cylinder $=\pi r^2h$

$=\pi(0.75)^2(1.2)$

$=0.675\pi\text{ m}^3$

3. Calculate the volume of the conical part and add to get total capacity.

Volume of cone $=\dfrac{1}{3}\pi r^2h$

$=\dfrac{1}{3}\pi(0.75)^2(0.5)$

$=0.09375\pi\text{ m}^3$

Total volume

$=0.675\pi+0.09375\pi$

$=0.76875\pi\text{ m}^3$

4. Convert the total volume from cubic metres to litres.

Since $1\text{ m}^3=1000$ litres,

$0.76875\pi\times1000=2415.71\ldots$

$\approx 2416\text{ litres}$

1. Convert the water volume to cubic metres and check that it fits inside the cone.

$500\text{ litres}=0.5\text{ m}^3$

Volume of full cone:
$\frac13\pi(0.75)^2(0.5)=0.09375\pi\approx0.2945\text{ m}^3$
Since $0.5>0.2945$, the cone is full and some water enters the cylinder.

2. Find the volume of water in the cylindrical part.

Volume above the cone:
$0.5-0.09375\pi\approx0.2055\text{ m}^3$

3. Find the height of water in the cylinder and add the cone height.

For the cylinder, base area $=\pi(0.75)^2=0.5625\pi$.

Height of water in cylinder:
$h=\frac{0.5-0.09375\pi}{0.5625\pi}\approx0.1163\text{ m}$

Total height of water from the bottom apex:
$0.5+0.1163=0.6163\text{ m}=61.6\text{ cm}$

1. Convert the given volume to cubic metres and compare it with the full cone volume.

$150\text{ litres}=0.150\text{ m}^3$

Full cone volume:
$\frac{1}{3}\pi(0.75)^2(0.5)=0.2945\ldots\text{ m}^3$

Since $0.150<0.2945\ldots$, the water is only in the conical part.

2. Use similarity of cones to find the water height from the apex.

Let the height of the water be $h$ m.

For similar cones,
$\left(\frac{h}{0.5}\right)^3=\frac{0.150}{0.2945\ldots}$

$h=0.5\sqrt[3]{\frac{0.150}{0.2945\ldots}}=0.358\ldots\text{ m}$

$0.358\ldots\text{ m}=35.8\ldots\text{ cm}$

1. Find the curved surface area of the cylindrical part.

Radius of mixer:

$r=\dfrac{1.5}{2}=0.75\text{ m}$

Curved surface area of cylinder:

$2\pi rh=2\pi(0.75)(1.2)=1.8\pi\text{ m}^2$

2. Find the curved surface area of the conical part and the total painted area for one mixer.

Slant height of cone:

$l=\sqrt{0.75^2+0.5^2}=\sqrt{0.8125}\text{ m}$

Curved surface area of cone:

$\pi rl=\pi(0.75)\sqrt{0.8125}$

Total area for one mixer:

$1.8\pi+0.75\pi\sqrt{0.8125}=7.779\ldots\text{ m}^2$

3. Find the number of cans needed for 6 mixers and round up.

Area for 6 mixers:

$6(7.779\ldots)=46.674\ldots\text{ m}^2$

Number of cans needed:

$\dfrac{46.674\ldots}{4.5}=10.37\ldots$

Since whole cans are needed, round up to $11$ cans.