Find the dimensions of the isosceles triangle of largest area that can be inscri

IMLOG10ct 2021-11-20 Answered
Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r.
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Expert Answer

Wasither1957
Answered 2021-11-21 Author has 17 answers

Step 1
From the figure, by Pythagoras theorem we can write
b2=r2x2
b=2r2x2(1)
image

Step 2
Area of Triangle =12 (Base)(Height)
A=12(b)(r+x)
Use Eqn(1) to substitute the value of b
A=12(2r2x2)(r+x)
A(x)=(r+x)r2x2
Step 3
Differentiate A(x)
A(x)=d[(r+x)r2x2]dx
Use Product Rule
A(x)=r2x2d[r+x]dx+(r+x)d[r2x2]dx
A(x)=r2x21+(r+x)d[r2x2]d(r2x2)×d(r2x2)dx
A(x)=r2x2+(r+x)12r2x2×(2x)
A(x)=r2x2rx+x2r2x2
Step 4
Solve for A(x)=0
r2x2rx+x2r2x2=0
Elizabeth Witte
Answered 2021-11-22 Author has 24 answers

Step 1
The equation of circle will be
x2+y2=r2
image

Step 2
Now, area of triangle
A=12(2x)(r+y)
=x(r+y)
A(y)=r2y2(r+y)
triangle derivative
A(y)=r2y2+(r+y)2y2r2y2
=r2ry2y2r2y2
equating to zero
r2ry2y2r2y2=0
r2ry2y2=0
y=r±9r24y=r or y=r2
rejecting, y=r
Therefore, we check to see y=r2 is maximum
A(y)>0 for y<r2 and A(y)<0 for y>r2
y=r2 is maximum
x2+(r2)2=r2x=32r
Dimension of triangle with maximum area is base
=2x=232r=3r
h=r+y=r+r2=32r
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