A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder top. The company will make 1545 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 , and round up your answer to the next square meter.

1 answer

To determine the total amount of aluminum needed to manufacture the mailboxes, we need to calculate the surface area of one mailbox. Each mailbox consists of a rectangular box with a half-cylinder on top.

Let’s begin by outlining the shapes involved:

  1. Rectangular Box: The rectangular part of the mailbox can be described by its height (h), width (w), and length (l).
  2. Half-Cylinder Top: The half-cylinder has a diameter (d) that relates to the width of the mailbox and a height that corresponds to the curve of the cylinder.

To calculate the surface area, we will follow these steps:

Step 1: Calculate the dimensions

Assume the typical dimensions for a mailbox:

  • Length of the box (l)
  • Width of the box (w)
  • Height of the box (h)
  • Diameter of the half-cylinder is equal to the width (d = w).
  • Height/length of the half-cylinder top (let's say it got a radius of r = d/2).

Step 2: Calculate the surface areas involved

Surface Area of the Rectangular Box:

The surface area \( A \) of a rectangular box without the top (since the cylinder sits on top) is given by:

\[ A_{box} = 2lh + 2lh + lw = 2lh + lw \]

Where:

  • Two sides are \( lw \) (left and right)
  • Two sides are \( lh \) (front and back)
  • One side is \( lw \) (bottom)

Surface Area of the Half-Cylinder Top:

The surface area of a cylinder is calculated using the formula \( 2\pi rh + 2\pi r^2 \). Since we only want the curved area of a half-cylinder, which also does not include the bottom, the area is:

\[ A_{cylinder} = \pi r h \] Where:

  • \( r = \frac{w}{2} \)
  • The height of the cylinder can be the same as the width w if we assume that it is a proper proportion.

Putting it together, the total surface area \( A_{total} \) for one mailbox would hence be:

\[ A_{total} = 2lh + lw + \frac{1}{2} \pi r h \]

Example:

Lets assume some dimensions:

  • Length (l) = 0.6 m
  • Width (w) = 0.3 m (thus diameter (d) = 0.3 m)
  • Height (h) = 0.2 m

Calculations:

  1. Calculate \( r \): \[ r = \frac{w}{2} = \frac{0.3}{2} = 0.15 , \text{m} \]

  2. Surface area of the box: \[ A_{box} = 2(0.6)(0.2) + (0.6)(0.3) = 0.24 + 0.18 = 0.42 , \text{m}^2 \]

  3. Curved surface area of half cylinder: \[ A_{cylinder} = \frac{1}{2} \pi (0.15)(0.3) \] Using \( \pi = 3.14 \): \[ A_{cylinder} = \frac{1}{2} \cdot 3.14 \cdot 0.15 \cdot 0.3 \approx \frac{1}{2} \cdot 3.14 \cdot 0.045 \approx 0.0353 \] So this total surface area \( A_{total} \) is: \[ A_{total} \approx 0.42 + 0.0353 \approx 0.4553 , \text{m}^2 \]

Step 3: Total for 1545 mailboxes

Now multiplying by the number of mailboxes: \[ A_{final} = 1545 \cdot 0.4553 \approx 703.75 , \text{m}^2 \]

Step 4: Round up the result

Rounding up gives us \( 704 , \text{m}^2 \) of aluminum needed.

Thus, the total square meters of aluminum required to make the 1545 mailboxes is:

\[ \boxed{704} , \text{m}^2 \]