To make a cylindrical tin bucket without cover, the bottom diameter is 4 decimeters, and the height is 5 decimeters. At least how large an area of tin is needed? To calculate the formula!

The bottom radius of the barrel is r = 4 / 2 = 2DM
The barrel height is h = 5DM
The base area is π R & # 178; = π × 2 & # 178; = 4 π (DM & # 178;)
The wall area is h × (2 π R) = 5 × (2 × π × 2) = 20 π (DM & # 178;)
Therefore, at least 24 π DM and 178; iron sheet is needed
The comprehensive formula is
(2πr)h+πr²
=(2×π×2)×5+π×2²
=20π+4π
=24π(dm²)

A cylindrical bucket has a bottom diameter of 30 cm and a height of 25 cm. How much iron sheet does it need to make such a bucket? What's the volume of this bucket?

3.14 × (30 △ 2) 2 + 3.14 × 30 × 25 = 3.14 × 225 + 2355 = 706.5 + 2355 = 3061.5 (square centimeter) ≈ 4 (square decimeter) 3.14 × (30 △ 2) 2 × 25 = 3.14 × 225 × 25 = 706.5 × 25 = 17662.5 (cubic centimeter) = 17662.5 (ML) a: making such a bucket needs at least 4 square decimeters of iron sheet, and the volume of the bucket is 17662.5 ml

This is a cylindrical tin bucket without a cover, 45 decimeters high and 9.42 decimeters in circumference at the bottom. How many square decimeters of tin can be used to make this bucket, and how many cubic decimeters of water can he hold at most

Radius = 9.42 △ 3.14 △ 2 = 1.5 decimeters
Iron sheet required = 9.42 × 45 + 3.14 × 1.5 × 1.5 = 430.965m2
Volume = 3.14 × 1.5 × 1.5 × 45 = 317.925 cubic decimeter
If you understand and solve your problem,

A cylindrical tin bucket without a cover is 5.1dm high and 18.84dm in circumference at the bottom. How many square decimeters does the bucket use at least?
How many kilograms of water can this bucket hold at most

Radius = 18.84 △ 2 △ 3.14 = 3 decimeters
Height = 5.1 decimeters
We need sheet iron
=Side area + bottom area
=2×3.14×3×5.1+3.14×3²
=124.344 square decimeters
It can hold water
=Volume × 1
=（3.14×3²×5.1）×1
=144.126 kg

To make a cylindrical tin bucket without cover, the bottom diameter is 4 decimeters, and the height is 5 decimeters. At least how large an area of tin is needed? To calculate the formula!