Question

Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side L if one side of the rectangle lies on the base of the triangle.

Answer 1

To find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle with side length \( L \), where one side of the rectangle lies on the base of the triangle, we start by establishing the configuration.

1. **Triangle Basics**: An equilateral triangle can be oriented such that one side (the base) lies along the x-axis, with its vertices at \( (0, 0) \), \( (L, 0) \), and \( \left( \frac{L}{2}, \frac{L \sqrt{3}}{2} \right) \).

2. **Rectangle Dimensions**: Let the rectangle have width \( w \) and height \( h \). The width of the rectangle will be \( w \) and will extend symmetrically from the center, which means the left corner is at \( \frac{L}{2} - \frac{w}{2} \) and the right corner is at \( \frac{L}{2} + \frac{w}{2} \).

3. **Rectangle Height**: The height of the rectangle, \( h \), can be calculated based on the position of the top corners of the rectangle within the triangle. The highest point of the triangle is at \( \left( \frac{L}{2}, \frac{L \sqrt{3}}{2} \right) \).

Using similar triangles, the height \( h \) of the rectangle at any point on the base can be expressed in terms of \( w \):
- The height of the triangle decreases linearly as we move horizontally. The slope of the triangle sides can be determined. The line from \( (0,0) \) to \( \left( \frac{L}{2}, \frac{L \sqrt{3}}{2} \right) \) has the equation:
\[
y = \sqrt{3} x
\]
- The height \( h \) at \( x = \frac{L}{2} \pm \frac{w}{2} \):
\[
h = \sqrt{3} \left( \frac{L}{2} - \frac{w}{2} \right)
\]

4. **Area of the Rectangle**: The area \( A \) of the rectangle can thus be expressed as:
\[
A = w \cdot h = w \cdot \sqrt{3} \left( \frac{L}{2} - \frac{w}{2} \right)
\]
Which simplifies to:
\[
A = \frac{\sqrt{3}}{2} w \left( L - w \right)
\]

5. **Maximizing the Area**: We want to maximize the function:
\[
A(w) = \frac{\sqrt{3}}{2} w (L - w)
\]
To find the maximum, we take the derivative of \( A \) with respect to \( w \) and set it to zero:
\[
\frac{dA}{dw} = \frac{\sqrt{3}}{2} (L - 2w)
\]
Setting this equal to zero gives:
\[
L - 2w = 0 \implies w = \frac{L}{2}
\]

6. **Finding height**: Substituting \( w = \frac{L}{2} \) back to find \( h \):
\[
h = \sqrt{3} \left( \frac{L}{2} - \frac{L}{4} \right) = \sqrt{3} \cdot \frac{L}{4} = \frac{L \sqrt{3}}{4}
\]

7. **Final Dimensions**: Thus, the dimensions of the rectangle of largest area inscribed in the triangle are:
- Width \( w = \frac{L}{2} \)
- Height \( h = \frac{L \sqrt{3}}{4} \)

The rectangle of largest area that can be inscribed in the equilateral triangle has dimensions:
\[
\text{Width: } \frac{L}{2}, \quad \text{Height: } \frac{L \sqrt{3}}{4}
\]