Maximum Area of Triangle  Problem with Solution
The first derivative is used to maximize the area of a triangle inscribed in a circle.
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 = (1 / 2) AB * CB

Using angle t, AB and CB may be expressed as follows
AB = AC * cos t and CB = AC * sin t

Substitute AC by AC * cos t and CB by AC * sin t in the above formula for the area, we obtain
S = (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 = (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.
dS / dt = (1/4)AC^{ 2} 2 cos (2t)

We now equate dS / dt to zero to find stationary points and interval of increase and decrease.
(1/4)AC^{ 2} 2 cos (2t) = 0

As point B slides along the circle, angle t changes fro 0 to 90 degrees. So the only solution to the above equation in the inteval (0 , 90) is
2 t = 90 or t = 45 degrees

As t changes from 0 to 45 , 2t changes from 0 to 90 and dS / dt is positive on this interval. As t changes from 45 to 90, 2t chnages from 90 to 180 degrees and 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 * cos (45 degrees) = 2 r sqrt(2)
and CB = AC * sin (45 degrees) = 2 r sqrt(2)

The area is maximum when t = 45 degrees which also means that the right triangle is isosceles.
More references on
calculus problems


Linear ProgrammingNew !
Online Step by Step Calculus Calculators and SolversNew !
Factor Quadratic Expressions  Step by Step CalculatorNew !
Step by Step Calculator to Find Domain of a Function New !
Free Trigonometry Questions with Answers

Interactive HTML5 Math Web Apps for Mobile LearningNew !

Free Online Graph Plotter for All Devices
Home Page 
HTML5 Math Applets for Mobile Learning 
Math Formulas for Mobile Learning 
Algebra Questions  Math Worksheets

Free Compass Math tests Practice
Free Practice for SAT, ACT Math tests

GRE practice

GMAT practice
Precalculus Tutorials 
Precalculus Questions and Problems

Precalculus Applets 
Equations, Systems and Inequalities

Online Calculators 
Graphing 
Trigonometry 
Trigonometry Worsheets

Geometry Tutorials 
Geometry Calculators 
Geometry Worksheets

Calculus Tutorials 
Calculus Questions 
Calculus Worksheets

Applied Math 
Antennas 
Math Software 
Elementary Statistics
High School Math 
Middle School Math 
Primary Math
Math Videos From Analyzemath
Author 
email
Updated: February 2015
Copyright © 2003  2015  All rights reserved