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### Maximum Area of Inscribed Triangle

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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.
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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).

Therefore:
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])

TrianMax.wri
12/14/2001

With Best Wishes,
Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Geometry
High School Triangles and Other Polygons

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