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It's been a while since I last tackled high-school math, and a friend asked me this question which I can't remember how to approach.

I have the following: $ y = -x^2 + 5x $ Which produces an inverse parabola, intersecting with the $x$ axis on $(0,0)$ and $(5,0)$ (I still remember that much).

The question is, place a point $a$ on the parabola graph, which will produce the largest perimeter rectangle with the axes.

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plot

The point is to find the $a$ where $P_{aboc}$ is the largest.

I would appreciate if anyone could point me to the right direction, even if not directly giving the solution!

Thanks in advance!

1 Answers 1

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As the point is on the parabola, write $\,A=(x,-x^2+5x)\,$ , so that the rectangle's perimeter is $H(x)=2(x+(-x^2+5x))=-2x^2+12x$

Proof 1: As $\,H\,$ is again an inverse parabola, its maximum value is obtained at its vertex $P\left(\frac{-b}{2a}\,,\,\frac{-\Delta}{4a}\right)=(3,16)\,,\,\mathrm{with}\,\,\,a=-2\,,\,b=12\,,\,c=0\,\,,\,\Delta:=b^2-4ac,$ so the point is $\,A=(3,6)\,$

Proof 2: Using differential calculus: $H'(x)=-4x+12=0\Longrightarrow x=3\Longrightarrow A=(3,6)$