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

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

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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)
$$A=\sqrt{2}r\sqrt{4r^2-(\sqrt{2}r)^2}$$
$$A=\sqrt{2}r\cdot\sqrt{4r^2-2r^2}$$
$$A=\sqrt{2}r\cdot\sqrt{2r^2}$$
$$A=\sqrt{2}r\cdot\sqrt{2}r$$
$$A=2r^2$$