Answer to Question #147884 in Trigonometry for yui

Question

(NOS. 2-5) PROBLEM 1. A cylindrical container of height equal to twice the diameter of its base can hold 12 liters (1L= 1,000 cm3) of water. Another cylindrical container with the same capacity has its height equal to three times the diameter of its base.



_______________________ 4) Determine the amount of aluminum required for making the first
container in cm2?


_____________________ 5) Determine the amount of aluminum required for making the
second container in cm2?

Answer 1

 The amount of aluminum required for making the first container in cm²

Solution

We, first of all calculate the diameter of the first container.

We know that the volume of a cylinder is given by, V=πr2h, where r = radius of the base surface.

Diameter = 2 × radius

Therefore, radius (r) = d​/2

In this case, height = twice the diameter of the base

Therefore, Volume = π (d/2)²​×h= 12 × 1000cm³

π (d2​)/4 × 2d=12×1000cm3, where π =22/7

We then solve for d

d = [(12000×2)/π]​^1/3 = 19.69cm​

Diameter = 19.69cm

To calculate, the amount of aluminum required, we assume that the containers are closed.

Amount of aluminum = total surface area = 2πrh + 2 πr²

Where radius = 19.69/2 = 9.85cm

Height = 19.69 × 2 = 39.38 cm.

Therefore, amount of aluminum for the first container = (2 × π × 9.85cm × 39.38cm) + [(2 × π × (9.85cm) ²] = 3048.04cm³

The amount of aluminum required for making the second container in cm²

Solution

Let d1​= the diameter of the base of the first container, h1​= the height of the first container and V1​= the volume of the first container.

Similarly, let d2​= the diameter of the base of the second container, h2​= the height of the second container and V2​= the volume of the second container.

Therefore, from the given information, h1​=2d1, V1​=π (d1​​/2)², h1​=1/2​π d13=1/16πh³1

h2=3d2, V1 = π (d2​​/2)², h2​= 3/4π d2³=1/36πh³2

We are given that, V1​=V2​=12L=12000cm³

​Therefore, 3/4π d2³=12,000cm³  

Calculate the value of d2

d2 = [(4 × 12000)/3π] ^1/3

d2 = 17.20cm

The diameter of second container = 17.20cm.

Therefore, radius of second container = 17.20/2 = 8.60cm

Height of second container = three times the diameter of the base = 3 × 17.20 = 51.60cm

Hence, the amount of aluminum required to make second container = total surface area = (2 × π × 8.60 × 51.60) + [2 × π × (8.6)²] = 3254.24cm²