(8.4 / 6) + 3.2
Hope this helps :)
The answer is either going to be the second one or the last one I'm going to go to with the last one if it's not the last one is the second one I hope I'm right if I'm not please let me know or if I am let me know
The area of the sector is 86.81π mm²
<u>Explanation:</u>
Given:
Radius of the sector, 25mm
Angle, α = 50°
Area of the sector, A = ?
We know:
Area of the sector = 
On substituting the value, we get:

Therefore, the area of the sector is 86.81π mm²
Answer:
Radius =6.518 feet
Height = 26.074 feet
Step-by-step explanation:
The Volume of the Solid formed = Volume of the two Hemisphere + Volume of the Cylinder
Volume of a Hemisphere 
Volume of a Cylinder 
Therefore:
The Volume of the Solid formed

Area of the Hemisphere =
Curved Surface Area of the Cylinder =
Total Surface Area=

Cost of the Hemispherical Ends = 2 X Cost of the surface area of the sides.
Therefore total Cost, C

Recall: 
Therefore:

The minimum cost occurs at the point where the derivative equals zero.


![-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518](https://tex.z-dn.net/?f=-27840%2B32%5Cpi%20r%5E3%3D0%5C%5C27840%3D32%5Cpi%20r%5E3%5C%5Cr%5E3%3D27840%20%5Cdiv%2032%5Cpi%3D276.9296%5C%5Cr%3D%5Csqrt%5B3%5D%7B276.9296%7D%20%3D6.518)
Recall:

Therefore, the dimensions that will minimize the cost are:
Radius =6.518 feet
Height = 26.074 feet