   Chapter 3.7, Problem 29E

Chapter
Section
Textbook Problem

Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r .

To determine

To find:

The dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r.

Explanation

1) Concept:

Apply the first derivative test for finding the dimensions of the isosceles triangle of the largest area that can be inscribed in a circle of radius r

2) Test:

First derivative test:

Suppose c is a critical number of a continuous function f defined on an interval.

(a) If f'(x)>0 for all x<c and f'(x)<0 for all x>c, then f(c) is the absolute maximum value of f.

(b) If f'(x)<0 for all x<c and f'(x)>0 for all x>c, then f(c) is the absolute minimum value of f.

3) Formula:

Area of the isosceles triangle is

A=bh2  where b and h are base and height of triangle

4) Given:

One side of the rectangle lies on the triangle

5) Calculation:

Let x be the distance between centre of circle and the base of the isosceles triangle is AB

From OAP

AB=2·AP

r2=AP2+x2

AP2=r2-x2

AP=r2-x2

Therefore,

AB=2r2-x2

Base of triangle has length 2r2-x2

Height of the triangle is (r+x)

Area of the isosceles triangle is given by,

A=bh2  where b and h are base and height of triangle

Therefore,A=bh2=2r2-x2·(r+x)2=r2-x2·r+x   where x[-r,r]

To find largest area that can be inscribed in a circle of radius r.

Ax=r2-x2·r+x (1)

Differentiating above equation w.r.to x,

A'x=ddxr2-x2·r+x

By using product rule for differentiation,

=ddxr2-x2·r+x

=r2-x2·ddxr+x+r+x·ddxr2-x2

=r2-x2·(1)+r+x·(-2x)2(r2-x2)

=r2-x2-xr+x(r2-x2)

=r2-x2-rx-<

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