Question

A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder top. The company will make 1823 mailboxes this week. If each mailbox has dimensions as shown in the figure below how many square meters of aluminum will be needed to make these mailboxes? In your calculations, use the value of 3.14 for pie, and round up your answer to the next square meter.

Answer 1

Answer:

$\large \boxed{\text{2145 m}^{2}}$

Step-by-step explanation:

The area of one mailbox = the area of a rectangular box - the top of the box plus half the area of a cylinder.

SA = SA(box) - SA(top) + ½SA(cylinder)

1. Surface area of box

The formula for the surface area of a rectangular box is

SA = 2(lw + lh + wh)

Data:  

 l =  0.55 m

w = 0.3    m

h =  0.4    m

Calculations:

 2(Top + Bottom = 2lw  = 2 × 0.55 × 0.3 = 0.33 m²

   2(Left + Right) = 2wh = 2 × 0.55 × 0.4 = 0.44 m²

 2(Front + Back) = 2lh   = 2 × 0.3   × 0.4 =  0.24 m²

                                                Total area =   1.01 m²

2. Surface area of cylinder

The  formula for the surface area of a cylinder is

SA = A(top) + A (base) + A(side)  = 2A(base) + A(side)

Data:

d =  0.3   m

h = 0.55 m

Calculations:

r = ½d  = ½ × 0.3 = 0.15 m

$\begin{array}{rcl}SA & = & 2\pi r^{2}+ 2\pi rh \\& = & 2 \times 3. 14\times 0.15^{2} +2\times 3. 14 \times 0.15\times 0.55\\& = & 6.28\times 0.0225 + 0.5181\\& = & 0.1413 + 0.5181\\& = & \mathbf{0.6594} \textbf{ m}^{\mathbf{2}}\\\end{array}$

3. Excluded area

         1 top = ½ × 0.33     m² = 0.165    m²

½ cylinder = ½ × 0.6594 m² = 0.3297 m²

                    Total excluded = 0.4947 m²

4. Surface area of 1 mailbox

SA = (1.01 + 0.6594 - 0.4927) m² = 1.1767 m²

5. Total area of 1823 mailboxes

$\text{Total area } = \text{1823 mailboxes} \times \dfrac{\text{1.1767 m}^{2}}{\text{1 mailbox}} = \textbf{2145 m}^{\mathbf{2}}\\\\\text{The company will have to use \large \boxed{\textbf{2145 m}^{\mathbf{2}}} of aluminium.}$