Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius .

he298c 2021-05-08 Answered
Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius .

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Expert Answer

likvau
Answered 2021-05-09 Author has 14352 answers

We apply Pythagoras Theorem.
The diagonal of this rectangle is \(r+r=2r\)
\(x^2+y^2=(2r)^2\)
\(x^2+y^2=4r^2\)
\(y^2=4r^2-x^2\)
\(y=\sqrt{4r^2-x^2}\)
Let it be A-area of rectangle.
\(A=xy\)
\(A=x\sqrt{4r^2-x^2}\)
Now, we differentiate A(x)
\(A'(x)=(x\sqrt{4r^2-x^2})'\)
\(=x'\sqrt{4r^2-x^2}+(\sqrt{4r^2-x^2})'x\)
\(=1\cdot\sqrt{4r^2-x^2}+(-\frac{x}{\sqrt{4r^2-x^2}})\)
\(=\sqrt{4r^2-x^2}-\frac{x^2}{\sqrt{4r^2-x^2}}\)
\(=\frac{4r^2-x^2-x^2}{\sqrt{4r^2-x^2}}\)
\(=\frac{4r^2-2x^2}{\sqrt{4r^2-x^2}}\)
Solve \(A'(x)=0\)
\(\frac{4r^2-2x^2}{\sqrt{4r^2-x^2}}=0\)
\(4r^2-2x^2=0\)
\(x=\sqrt{2}r,x=-\sqrt{2}r\)
x must be positive, so \(x=\sqrt{2}r\)
Largest area is: (subtitute \(x=\sqrt{2}r\) in equation of area)
\(A=\sqrt{2}r\sqrt{4r^2-(\sqrt{2}r)^2}\)
\(A=\sqrt{2}r\cdot\sqrt{4r^2-2r^2}\)
\(A=\sqrt{2}r\cdot\sqrt{2r^2}\)
\(A=\sqrt{2}r\cdot\sqrt{2}r\)
\(A=2r^2\)

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