Question

a company manufactures aluminum mailboxes in the shape of a box with a half - cylinder top. the company will make 1485 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 3.14 for π, and round up your answer to the next square meter. your answer

Explanation:

Step1: Calculate the surface area of the rectangular part

The rectangular part has three different faces.

• Two faces with dimensions \(0.7\times0.4\): Area \(= 2\times(0.7\times0.4)=2\times0.28 = 0.56\)
• Two faces with dimensions \(0.7\times0.55\): Area \(=2\times(0.7\times0.55)=2\times0.385 = 0.77\)
• One face with dimensions \(0.4\times0.55\): Area \(=0.4\times0.55 = 0.22\)

Step2: Calculate the surface area of the half - cylinder part

The formula for the lateral surface area of a full - cylinder is \(A = 2\pi rh\). For a half - cylinder, it is \(A=\pi rh\), where \(r=\frac{0.4}{2}=0.2\)m and \(h = 0.7\)m.
The area of the half - cylinder's lateral surface \(A_{lateral}=\pi\times0.2\times0.7=3.14\times0.2\times0.7 = 0.4396\)
The area of the circular end (since it's a half - cylinder, we consider one - half of a full - circle's area. But in the context of the mailbox, if we assume no top or bottom for the half - cylinder part in terms of the aluminum sheet used for the outer surface, we mainly focus on the lateral part. However, if we consider the flat side of the half - cylinder (the rectangular part that closes the half - cylinder), its area is \(0.4\times0.7=0.28\)

Step3: Sum up the areas for one mailbox

Total surface area of one mailbox \(S=(0.56 + 0.77+0.22)+(0.4396 + 0.28)\)
\(S=(1.55)+(0.7196)=2.2696\)

Step4: Calculate the total area for 1485 mailboxes

Total area \(A = 1485\times2.2696\)
\(A=1485\times2.2696=1485\times(2 + 0.2+0.06+0.0096)\)
\(A=(1485\times2)+(1485\times0.2)+(1485\times0.06)+(1485\times0.0096)\)
\(A = 2970+297+89.1+14.256\)
\(A=3370.356\approx3370.4\)

Answer:

\(3370.4\)