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Maximum Area of Inscribed Triangle

Date: 12/10/2001 at 20:52:23
From: Shawn Vaillancourt
Subject: Calculus - Max/Mins for Trig

An isosceles triangle is inscribed in a circle of radius R. Find the 
value of Theta that maximizes the area of the triangle. 

There is a diagram included that shows the isoceles triangle in the 
circle, but theta (the angle between the two equal sides) is marked as 
2theta. I figure with common sense that the angle should be 60 degrees 
or pi/3 radians, but I don't know how to do it the calculus way.

Date: 12/14/2001 at 12:06:03
From: Doctor Fwg
Subject: Re: Calculus - Max/Min (Max. Area of Inscribed Triangle) 

Dear Shawn,

Here is help with a solution to finding angles of a triangle having 
maximum area if it is drawn within a circle so that the triangle's 
vertices just touch the circle's perimeter (i.e., a triangle inscribed 
within a circle).

First draw an isosceles triangle within a circle having an arbitrary 
radius equal to R:


The angle at the top of the triangle is assigned to Phi. One half of 
this angle is Phi/2. The two equal angles at the base of the triangle 
are assigned to Theta. The two equal side lengths of this triangle are 
assigned to L. The base of the triangle is assigned to B. The height 
of the triangle is assigned to H.

Now, it is well known that the Area (A) of any triangle is:  
A = (1/2)BH.

To find the angles associated with the inscribed triangle (described 
above) having a maximum area, one may first re-express the area 
formula above in terms involving only Phi (or Phi/2) and R.

You should be able to show that:
L/2 = Rcos(Phi/2), so
L = 2Rcos(Phi/2), and
B = 2[2Rcos(Phi/2)sin(Phi/2)] = 4Rcos(Phi/2)sin(Phi/2), and
H = 2Rcos(Phi/2)cos(Phi/2) = 2Rcos^2(Phi/2).

A = (1/2)BH = (1/2)[4Rcos(Phi/2)sin(Phi/2)][2Rcos^2(Phi/2)], or
A = 4R^2[cos^3(Phi/2)sin(Phi/2)] .

Now, one may take the first derivative of A with respect to Phi [i.e., 
dA/dPhi] by holding R constant because the angles associated with the 
inscribed triangle having a maximum area are independent of the 
circle's size. If [dA/dPhi] is set to equal zero, one may solve for 
the value of Phi that satisfies this condition. Once that value of Phi 
is known, one may solve for Theta because: Phi + 2Theta = 180. Then, 
all the angles associated with the triangle having a maximum area will 
be found.

Upon simplification, one should find that:
[dA/dPhi] = 2R^2cos^2(Phi/2)[cos^2(Phi/2) - 3sin^2(Phi/2)].

Setting: [cos^2(Phi/2) - 3sin^2(Phi/2)] = 0, one can show that:
Phi/2 = 30 deg, so:
Phi = 60 deg, therefore,
2Theta = 120 deg, and,
Theta = 60 deg.

Therefore, the inscribed triangle having a maximum area is an 
equilateral triangle.

The tricky part of this problem is in finding values of B and H only 
in terms of R and Phi. I hope this outline will be of help to you.

(Soln. dedicated to our beloved son Jim [b:12/14/69 - d:10/18/95])


With Best Wishes,  
Doctor Fwg, The Math Forum   
Associated Topics:
High School Calculus
High School Geometry
High School Triangles and Other Polygons

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