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For a triangle with a given base and altitude, an isosceles triangle has the minimum perimeter.
At $x=0.0$, $P'(x) = -\dfrac{1}{\sqrt{1+Y^{2}}}$
At $x=0.5$, $P'(x) = 0$
At $x=1.0$, $P'(x) = +\dfrac{1}{\sqrt{1+Y^{2}}}$
At $x=0.5$, $P'(x) = 0$
At $x=1.0$, $P'(x) = +\dfrac{1}{\sqrt{1+Y^{2}}}$
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