# Maximum Area of Triangle - Optimization Problem with Solution

The first derivative is used to maximize the area of a triangle inscribed in a circle. An optimization problem with solution.

## 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 length of sides AB and CB so that the area of triangle ABC is maximum.__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 \[ AB = AC \times \cos t \quad \text{and} \quad CB = AC \times \sin t \] Substitute \(AC\) by \(AC \times \cos t\) and \(CB\) by \(AC \times \sin t\) in the above formula for the area, we obtain \[ S = \frac{1}{2} AC^2 \cos t \sin t \] We now use the trigonometric formula \(\sin(2t) = 2 \sin t \cos t\) to express the area \(S\) as follows. \[ S = \frac{1}{4}AC^2 \sin(2t) \] Since the radius \(r\) is constant, the length of the diameter \(AC\) 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 need to find the first derivative and the stationary points. \[ \frac{dS}{dt} = \frac{1}{4}AC^2 2 \cos(2t) \] We now equate \(\frac{dS}{dt}\) to zero to find stationary points and interval of increase and decrease. \[ \frac{1}{4}AC^2 2 \cos(2t) = 0 \] As point B slides along the circle, angle \(t\) changes from 0 to 90 degrees. So the only solution to the above equation in the interval \( (0 , 90) \) is \[ 2t = 90 \quad \text{or} \quad t = 45 \text{ degrees} \] As \(t\) changes from 0 to 45 , \(2t\) changes from 0 to 90 and \(\frac{dS}{dt}\) is positive on this interval. As \(t\) changes from 45 to 90, \(2t\) changes from 90 to 180 degrees and \(\frac{dS}{dt}\) is negative on this interval. \(t = 45\) degrees is the location of a maximum value for the area \(S\).

When \(t = 45\) degrees, the area of the inscribed right triangle is maximum. The length of sides \(AB\) and \(CB\) are given by \[ AB = AC \times \cos (45 \text{ degrees}) = 2r \sqrt{2} \] and \[ CB = AC \times \sin (45 \text{ degrees}) = 2r \sqrt{2} \] The area is maximum when \(t = 45\) degrees which also means that the right triangle is isosceles.