BuyFind

Single Variable Calculus: Concepts...

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

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

Answer to Problem 41RE

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 .

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!