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A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder top. The company will make 1728 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 X, and round up your answer to the next square meter.?

A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder-example-1
User Siebe
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Answer:

1759 square meters

Explanation:

You want the surface area of a cuboid with a half-cylinder top.

Lateral area

The lateral area of the figure is the product of the length of the mailbox (0.45 m) and the perimeter of the end. The perimeter of the end is the sum of the lengths of the straight sides and half the circumference of a circle with diameter 0.3 m.

P = 0.3 + 2·0.4 + π/2(0.3) = 1.571 . . . . . meters

LA = Ph = (1.571 m)(0.45 m) = 0.70695 m²

End area

The end area is twice the area of the rectangular portion of the end, plus the area of a circle 0.3 m in diameter.

EA = (0.3 m)(0.4 m) + 3.14(0.3/2 m)² = 0.31065 m²

Total area

The total area of 1 mailbox is ...

LA +EA = 0.70695 m² +0.31065 m² = 1.0176 m²

Then the area of 1728 mailboxes is ...

1728 × 1.0176 m² ≈ 1758.4 m² ≈ 1759 m²

About 1759 square meters of aluminum will be needed for the 1728 mailboxes.

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Additional comment

This presumes there is no waste in cutting the semicircular shape from the supplied aluminum.

A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder-example-1
User Surfmuggle
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