Matematické Fórum

stuart clark · 01. 08. 2014 08:09

Calculation of Maximum value of $f(x) = x\cdot \left(\frac{\sqrt{100-x^2}+\sqrt{81-x^2}}{2}\right)$

Xellos · 01. 08. 2014 15:46

It's clear that $f(x)$ has the same sign as $x$ , so we only need positive $x$ .

$f(x)=\frac{19x}{2\left(\sqrt{100-x^2}-\sqrt{81-x^2}\right)}=\frac{19}{2}\left(\sqrt{\frac{100}{x^2}-1}-\sqrt{\frac{81}{x^2}-1}\right)^{-1}$

$\left(\sqrt{\frac{100}{x^2}-1}-\sqrt{\frac{81}{x^2}-1}\right)'=\left(\frac{-100}{x^3\sqrt{\frac{100}{x^2}-1}}+\frac{81}{x^3\sqrt{\frac{81}{x^2}-1}}\right)=0$

for extremes. From it, we find

$\sqrt{81-x^2}=\frac{81}{100}\sqrt{100-x^2}$
$x^2\left(1-\frac{81^2}{100^2}\right)=81-\frac{81^2}{100}$
$x=\sqrt\frac{100\cdot81}{100+81}$

for which $f(x)=45$ . In general, we can derive this way that the maximum of

$f(x)=x\frac{\sqrt{a^2-x^2}+\sqrt{b^2-x^2}}{2}$

is $\frac{ab}{2}$ .