Question

A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder top. The company will make 1863 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 П, and round up your answer to the next square meter.

Answer 1

Answer:

3433 m²

Step-by-step explanation:

From the image, we have a rectangular box without cover and half a cylinder on top.

Formula for surface area of rectangular box with top is;

S = 2(lh + wh + lw)

From the image,

l = 0.6 m

w = 0.4 m

h = 0.55 m

Thus;

S = 2((0.6 × 0.55) + (0.4 × 0.55) + (0.6 × 0.4))

S = 1.58 m²

Now, since the top is not included for this figure, then;

Surface area of this rectangular box is;

S1 = 1.58 - (lw) = 1.58 - (0.4 × 0.6) = 1.34 m²

Surface area of a cylinder is;

S = 2πr² + 2πrh

r is radius and in this case = 0.4/2 = 0.2 m

h = 0.6

S = 2π(0.2² + (0.2 × 0.6))

S = 1.005 m²

Since it is half cylinder, then we have;

S2 = 1.005/2

S2 = 0.5025 m²

Total surface area; S_t = S1 + S2

S_t = 1.34 + 0.5025

S_t = 1.8425 m²

This is the surface area of one mail box.

Thus, for 1863 mailboxes, total surface area is;

S = 1863 × 1.8425 = 3432.5775 m²

Approximating to the nearest Sq.m gives;

S = 3433 m²