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 pi, and round up your answer to the next square meter.

Answer 1

Answer:$8746.65\ m^2$

Step-by-step explanation:

Given

Base of mail box is $0.4\ m\times 0.7\ m$

height of mail box $h=0.55\ m$

Total area required for one mail box

$A_t=\text{surface area of cubiod} - \text{Top of cubiod +Semi-cylinder area}$

$A_t=2(lb+bh+hl)-lb+\pi rh'$

where $b=0.4\ m$

$l=0.7\ m$

$h=0.55\ m$

$r=\frac{0.4}{2}=0.2\ m$

$h'=0.7\ m$

Substituting values

$A_t=2(0.7\times 0.4+0.4\times 0.55+0.55\times 0.7)-0.4\times 0.7+\pi \times 0.2\times 0.7$

$A_t=2(0.28+0.385+0.22)-0.28+\pi \times 0.14$

$A_t=1.77-0.28+4.4$

$A_t=5.89\ m^2$

Therefore for 1485 mailboxes Area required

$A_o=1485\times 5.89=8746.65\ m^2$