How do you find the largest possible area for a rectangle inscribed in a circle of radius 4?

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How do you find the largest possible area for a rectangle inscribed in a circle of radius 4?

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Answer 1 · 2016-09-18T10:57:34

$32$ $32$ units of area

Explanation:

An inscribed rectangle has diagonals of length $2 r$ $2r$ and has sides $(a, b)$ $(a,b)$ measuring $0 < a < 2 r$ $0< a < 2r$ , $b = \sqrt{{(2 r)}^{2} - a^{2}}$ $b = sqrt((2r)^2-a^2)$ so the rectangle area is

$A = a \sqrt{{(2 r)}^{2} - a^{2}}$ $A = a sqrt((2r)^2-a^2)$ for $0 < a < 2 r$ $0< a < 2r$ .

Its maximum occurs at $a_{0}$ $a_0$ such that

${(\frac{d A}{d a})}_{a_{0}} = 0$ $((dA)/(da))_(a_0) = 0$ or

$\frac{2 (a_{0}^{2} - 2 r^{2})}{\sqrt{4 r^{2} - a_{0}^{2}}} = 0$ $(2 (a_0^2 - 2 r^2))/sqrt[4 r^2-a_0^2] = 0$ giving

$a_{0} = \sqrt{2} r$ $a_0 = sqrt(2)r$ and at this value

$A_{0} = 2 r^{2} = 2 \times 4^{2} = 32$ $A_0 = 2r^2=2xx4^2= 32$

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How do you find the largest possible area for a rectangle inscribed in a circle of radius 4?

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