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:
- Rectangular Box: The rectangular part of the mailbox can be described by its height (h), width (w), and length (l).
- 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:
-
Calculate \( r \): \[ r = \frac{w}{2} = \frac{0.3}{2} = 0.15 , \text{m} \]
-
Surface area of the box: \[ A_{box} = 2(0.6)(0.2) + (0.6)(0.3) = 0.24 + 0.18 = 0.42 , \text{m}^2 \]
-
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:
\[ \boxed{704} , \text{m}^2 \]
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