The first derivative is used to maximize the area of a triangle inscribed in a circle. An optimization problem with detailed solution is presented below.
In what follows, \( C \) represents the constant of integration where applicable.
Problem
In the picture below, triangle ABC is inscribed inside a circle of center O and radius \( r \). For a constant radius \( r \) of the circle, point B slides along the circle so that the area of ABC changes. Find the lengths of sides AB and CB so that the area of triangle ABC is maximum.
Show Solution to Problem
Since the center O of the circle is on the side AC of the triangle, AC is a diameter of the circle and triangle ABC is a right triangle (Thales's theorem). Its right angle is at B and its hypotenuse AC has length equal to \( 2r \) (see figure below).
The area \( S \) of the triangle is given by:
\[ S = \frac{1}{2} AB \times CB \]Using angle \( t \), \( AB \) and \( CB \) may be expressed as follows:
\[ AB = AC \cos t \quad \text{and} \quad CB = AC \sin t \]Substitute \( AB \) and \( CB \) into the formula for the area:
\[ S = \frac{1}{2} AC^2 \cos t \sin t \]Using the trigonometric identity \( \sin(2t) = 2 \sin t \cos t \), we express the area \( S \) as:
\[ S = \frac{1}{4} AC^2 \sin(2t) \]Since the radius \( r \) is constant, the length of the diameter \( AC = 2r \) is also constant. Hence, the area depends on angle \( t \) only as point B slides along the circle. To find \( t \) so that \( S \) is maximum, we calculate the first derivative with respect to \( t \):
\[ \frac{dS}{dt} = \frac{1}{4} AC^2 \cdot 2 \cos(2t) = \frac{1}{2} AC^2 \cos(2t) \]Equating \( \frac{dS}{dt} \) to zero to find stationary points:
\[ \frac{1}{2} AC^2 \cos(2t) = 0 \implies \cos(2t) = 0 \]As point B slides along the circle, angle \( t \) changes from \( 0 \) to \( 90^\circ \). The solution to \( \cos(2t) = 0 \) in the interval \( (0^\circ, 90^\circ) \) is:
\[ 2t = 90^\circ \implies t = 45^\circ \]As \( t \) changes from \( 0^\circ \) to \( 45^\circ \), \( 2t \) changes from \( 0^\circ \) to \( 90^\circ \) and \( \frac{dS}{dt} \) is positive. As \( t \) changes from \( 45^\circ \) to \( 90^\circ \), \( 2t \) changes from \( 90^\circ \) to \( 180^\circ \) and \( \frac{dS}{dt} \) is negative. Therefore, \( t = 45^\circ \) gives the maximum value for the area \( S \).
When \( t = 45^\circ \), the lengths of sides \( AB \) and \( CB \) are:
\[ AB = AC \cos(45^\circ) = 2r \left(\frac{\sqrt{2}}{2}\right) = r\sqrt{2} \] \[ CB = AC \sin(45^\circ) = 2r \left(\frac{\sqrt{2}}{2}\right) = r\sqrt{2} \]Conclusion: The area is maximum when \( t = 45^\circ \), which means the inscribed right triangle is an isosceles right triangle with side lengths \( AB = CB = r\sqrt{2} \).