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.

triangle for problem 1

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)

    triangle for problem 1

  • 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

Step by Step Math Worksheets SolversNew !
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 - e-mail

Updated: February 2015

Copyright 2003 - 2016 - All rights reserved