Get the answer to "Find the measurements of the greatest area possible..."

he298c

Answered question

2021-05-08

Find the measurements of the greatest area possible rectangle that can be enclosed in a circle.

Answer & Explanation

likvau

Skilled2021-05-09Added 75 answers

We apply Pythagoras Theorem.
The diagonal of this rectangle is $r + r = 2 r$
$x^{2} + y^{2} = (2 r)^{2}$
$x^{2} + y^{2} = 4 r^{2}$
$y^{2} = 4 r^{2} - x^{2}$
$y = \sqrt{4 r^{2} - x^{2}}$
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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