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.

Short answer

The dimensions of the rectangle are one side \(a = \frac{L}{2}\) and the other side \(b = \frac{{\sqrt 3 L}}{4}\).

Step-by-step solution

To sketch the figure

Since the rectangle is inscribed in the equilateral triangle, the base of the triangle and the side of the rectangle which lies on the base must have the same midpoint.

 To find the dimensions of the rectangle

Let\(a > 0\)be the length of one side of the rectangle which lies on the base of the triangle.

Let\(b > 0\)be the length of other side of the rectangle.

Therefore, the area of the rectangle is,

\(A = a \cdot b\,\,\, \cdots \cdots \left( 1 \right)\).

From figure in step 1, consider the triangle at the bottom on the right side, we have,

\(\begin{array}{l}\tan \left( {60^\circ } \right) = \frac{b}{{\frac{{L - a}}{2}}}\\ \Rightarrow \sqrt 3 = \frac{{2b}}{{L - a}}\\ \Rightarrow b = \frac{{\sqrt 3 }}{2}\left( {L - a} \right)\end{array}\)

Thus, equation \(\left( 1 \right)\) becomes,

\(\begin{array}{l}A\left( a \right) = a \cdot \left( {\frac{{\sqrt 3 }}{2}\left( {L - a} \right)} \right)\\ \Rightarrow A\left( a \right) = \frac{{\sqrt 3 }}{2}La - {a^2}\end{array}\)

In order to find the length of a side of a rectangle with maximum area, we need the derivative of the area function.

Therefore,\(A'\left( a \right) = \frac{{\sqrt 3 }}{2}\left( {L - 2a} \right)\).

Thus, the only critical number for\(A\)is,\(a = \frac{L}{2}\).

Here,\(A'\left( a \right) < 0,\,\,\forall a > \frac{L}{2}\)\(\)and\(A'\left( a \right) > 0,\,\,\forall a < \frac{L}{2}\).

Hence,\(A\)has an absolute maximum at\(a = \frac{L}{2}\).

Thus, the area of the inscribed rectangle is the greatest when the lengths of the sides are\(a = \frac{L}{2}\) and

\(\begin{array}{l}b = \frac{{\sqrt 3 }}{2}\left( {L - \frac{L}{2}} \right)\\ \Rightarrow b = \frac{{\sqrt 3 L}}{4}\end{array}\)