"Find the dimensions of the rectangle of largest..." was answered by our community

Clifland

Answered question

2021-08-13

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

Layton

Skilled2021-08-14Added 89 answers

We apply Pythagoras Theorem.
The diagonal of this rectangle is r+r=2r

x^{2} + y^{2} = {(2 r)}^{2}

x^{2} + y^{2} = 4 r^{2}

y^{2} = 4 r^{2} - x^{2}

y = \sqrt{4 r^{2} - x^{2}}

Let it be A-area of rectangle.

A = x y

A = x \sqrt{4 r^{2} - x^{2}}

Now, we differentiate A(x)

A^{'} (x) = (x {\sqrt{4 r^{2} - x^{2}}}^{'}

= x^{'} \sqrt{4 r^{2} - x^{2}} + {(\sqrt{4 r^{2} - x^{2}})}^{'} x

= 1 \cdot \sqrt{4 r^{2} - x^{2}} + (- \frac{x}{\sqrt{4 r^{2} - x^{2}}})

= \sqrt{4 r^{2} - x^{2}} - \frac{x^{2}}{\sqrt{4 r^{2} - x^{2}}}

= \frac{4 r^{2} - x^{2} - x^{2}}{\sqrt{4 r^{2} - x^{2}}}

= \frac{4 r^{2} - 2 x^{2}}{\sqrt{4 r^{2} - x^{2}}}

Solve A'(x)=0

\frac{4 r^{2} - 2 x^{2}}{\sqrt{4 r^{2} - x^{2}}} = 0

4 r^{2} - 2 x^{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{4 r^{2} - {(\sqrt{2} r)}^{2}}

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

A = \sqrt{2} r \cdot \sqrt{2 r^{2}}

A = \sqrt{2} r \cdot \sqrt{2} r

A = 2 r^{2}

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