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

Answer 1

9514 1404 393

Answer:

  3433 m^2

Step-by-step explanation:

The total surface area is the sum of the end areas and the lateral area.

The end areas together are a circle of diameter 0.4 m and two rectangles that are 0.4 m by 0.55 m.

  End areas = πr^2 + 2LW = 3.14(0.2 m)^2 + 2(0.4 m)(0.55 m)

  = 0.1256 m^2 +0.44 m^2 = 0.5656 m^2

The perimeter of the end piece is half the circumference of a 0.4 m circle and three sides of a 0.4 m by 0.55 m rectangle.

  P = (1/2)πd + 2L +W = 1/2·3.14·0.4 m + 2(0.55 m) + 0.4 m = 2.128 m

The lateral area of the mailbox is the product of its length and the perimeter of the end piece:

  LA = PL = (2.128 m)(0.6 m) = 1.2768 m^2

Then the total surface area of the mailbox is ...

  0.5656 m^2 + 1.2768 m^2 = 1.8424 m^2

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The area of aluminum required for 1863 mailboxes is ...

  1863 × 1.8424 m^2 ≈ 3433 m^2

About 3433 square meters of aluminum are needed to make these mailboxes.