Eric Nicholas K's answer to Kai's Secondary 4 A Maths Singapore question.
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We need to define the length of the rectangle and the breadth of the rectangle in variables x and y. The area will be length times breadth, which we will eventually express in terms of x only.
The rest is a maxima/minima question. I am lazy to prove that the value is indeed a maximum.
The rest is a maxima/minima question. I am lazy to prove that the value is indeed a maximum.
Date Posted:
5 years ago
4(a² - x²)^(-½) > 0 for any real value of a and x. So when dA/dx = 0 , a² - 2x² = 0
So the stationary point is at x = √2 a /2
if x > √2 a /2, then your 2x² > a².
So a² - 2x² < 0 → dA/dx < 0
if x < √2 a /2, then your 2x² < a².
So a² - 2x² > 0 → dA/dx > 0
Since dA/dx is negative beyond the stationary point, and positive for values lower than the stationary point, x = √2 a /2 must be a maximum point.
(We only consider values of x > 0 since length must be positive and non zero)
Also good to note for any inscribed rectangle in the circle, the diagonal length is always equal to 2a since it is always the diameter of the circle. So the hypotenuse is always a constant
So the stationary point is at x = √2 a /2
if x > √2 a /2, then your 2x² > a².
So a² - 2x² < 0 → dA/dx < 0
if x < √2 a /2, then your 2x² < a².
So a² - 2x² > 0 → dA/dx > 0
Since dA/dx is negative beyond the stationary point, and positive for values lower than the stationary point, x = √2 a /2 must be a maximum point.
(We only consider values of x > 0 since length must be positive and non zero)
Also good to note for any inscribed rectangle in the circle, the diagonal length is always equal to 2a since it is always the diameter of the circle. So the hypotenuse is always a constant
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