Question

Sam and Carlos has decided to build a cylindrical water tank made out of polyethylene accompanied by a rectangular prism base made out of concrete. Given: Density of concrete = 2.4g/cm3Density of polyethylene = 0.900g/cm3The water tank has a height of 8ft and a width of 5ft. The concrete base has a length of 9ft, a width of 7ft, and a height of 1 foot. Calculate the amount of material needed to build each object.Calculate the cost with an explanation or spreadsheet to show the breakdown of costs. *Note that the costs you come up with should be a reasonable price and should altogether add up to $20,000 or less.

Answer 1

Explanation

From the statement, we have the following objects.

1) A water tank made of polyethylene with:

• height h = 8 ft,

,

• width or diameter d = 5 ft,

,

• thickness t.

2) A concrete base with:

• base b = 9 ft,

,

• width w = 7 ft,

,

• height h = 1 ft.

We also know that:

• the density of polyethylene is ρₚ = 0.900 g/cm³,

,

• the density of concrete is ρc = 2.4 g/cm³.

We convert the densities to g/ft³:

$\begin{gathered} \rho_p=0.900*\frac{g}{\text{ cm}^3}=0.900*\frac{g}{\text{ cm}^3}*(\frac{30.48\text{ cm}}{1\text{ ft}})^3\cong25485.16\frac{g}{\text{ ft}^3}\cong25.49\frac{kg}{ft^3}, \\ \rho_c=2.4*\frac{g}{\text{ cm}^3}=2.4*\frac{g}{\text{ cm}^3}*(\frac{30.48\text{ cm}}{1\text{ ft}})^3\cong67960.43\frac{g}{\text{ ft}^3}\cong67.96\frac{kg}{ft^3}. \end{gathered}$

1) Volume of the water tank

Using the data from the water tank, we make the following diagram:

The volume of the water thank is given by:

$V_T=t\cdot A_b+t\cdot A_s=t\cdot(A_b+A_c).$

Where:

• Ab = area of the base,

• Ac = area of the curved side,

,

• t = thickness of the tank.

The area of the base is:

$A_b=\frac{\pi d^2}{4}\cong\frac{3.14\cdot(5ft)^2}{4}\cong19.63\text{ ft}^2.$

The area of the curved side is:

$A_c=l\cdot h=(\pi\cdot d)\cdot h\cong3.14\cdot5\text{ ft}\cdot8\text{ ft}=125.6\text{ ft}^2.$

Replacing th values of Ab and Ac in the formula above, we get thevolume of the tank

$V_T=t\cdot(19.63\text{ ft}^2+125.6\text{ ft}^2)=t\cdot145.23\text{ ft}^2.$

The mass of the tank is given by:

$m_T=V_T\cdot\rho_p\cong t\cdot142.23ft^2\cdot25.49\frac{kg}{\text{ ft}^3}=3625.44\text{ kg}\cdot\frac{t}{\text{ ft}}.$

By replacing the value of the thickness, we get the mass of the tank.

2) Voume of the base

The concrete base is a rectangular prism. Its volume is given by:

$V_b=b\cdot w\cdot h.$

Replacing the values of b, w and h, we get:

$V_b=9ft\cdot7ft\cdot1ft=63\text{ ft}^3.$

The mass of the base is:

$m_b=V_b\cdot\rho_b=63\text{ ft}^3\cdot67.96\frac{kg}{\text{ ft}^3}=4281.48\text{ }kg$Answer

• The ,mass of the water tank, is:

$m_T=V_T\cdot\rho_p\cong t\cdot142.23ft^2\cdot25.49\frac{kg}{\text{ ft}^3}=3625.44\text{ kg}\cdot\frac{t}{\text{ ft}}$

Replacing the value of the thickness t (in ft), we get the mass.

• The ,mass of the concrete base, is:

$m_b=V_b\cdot\rho_b=63\text{ ft}^3\cdot67.96\frac{kg}{\text{ ft}^3}=4281.48\text{ }kg$