# Tofind: thesmallest possible area of an isosceles that can be circumscribed in a circle. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4, Problem 41RE
To determine

## Tofind:thesmallest possible area of an isosceles that can be circumscribed in a circle.

Expert Solution

The minimum area is 33r2 .

### Explanation of Solution

Given:

Radius of circle is r .

Concept used:

Area of triangle =12×base×height .

Trigonometric formula:

tan2θ=2tanθ1-tan2θ .

If d2ydx2>0. the concave will be open upward and local minima can be found.

If d2ydx2<0. the concave will open downward and local maxima can be found.

Calculation:

By making a diagram of the circle inscribed in the isosceles triangle

The radius of circle r will tends to perpendicular and base will be the half of base of the isosceles triangle let it be denoted by d.

Through this it can be solved as:

tanθ=rd or tan2θ=hd .

dtan2θ=h .

Area of triangle =12×base×height .

12×(2d)×h=d×h=d×dtan2θ .

Using trigonometric formula:

tan2θ=2tanθ1-tan2θ .

Area= 2d2tanθ1-tan2θ .

Substituting tanθ=rd .

2d2(rd)1-(rd)2 .

2rd3d2r2 .

A(d)=2rd3d2r2 .

Differentiating with respect to d to get:

A(d)=(d2-r2)(6rd2)-(2rd3)(2d)(d2-r2) .

A(d)=2rd2(d2-r2)(d2-3r2)

A(d)=2rd2(d2-r2)(d+3r)(d-3r) .

When d<3r,A(d) is decreasing.

When d>3r,A(d) is increasing.

Therefor the d=3r is Minima.

A(3r)=2r(3r)3(3r)2-r2 .

63r43r2-r2 = 63r42r2 = 33r2 .

Hence the minimum area is 33r2 .

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