A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the maximum length of the hypotenuse is `(a^(2/3)+b^(2/3))^(3/2)`.

Explanation

Let a point P on the hypotenuse AC of a right `Delta ABC` such that
` PL bot AB = a and PM bot BC = b` Again, let
` angleAPL = angleACB = theta` then,
`Ap = a sec theta, PC = b cosec theta`
Again if the length of hypotenuse is l, then `l = AP + PC rArr l = a sec theta + b cosec theta, 0 lt theta lt pi/2`
Differentiate w.r.t.` theta`,
`(dl)/(d theta) = a sec theta tan theta- b cosec theta cot theta`
For maxima/minima, ` (dl)/(d theta) = 0`
` rArr a sec theta tan theta = b cosec theta cot theta`
`rArr (a sin theta)/(cos^(2) theta) = (b cos theta)/(sin^(2) theta)rArr (sin^(3)theta)/(cos^(3)theta) = b/a`
` rArrr tan^(3)theta=b/a rArr tan theta = (b/a)^(1/3)`
Now `(d^(2)l)/(d theta^(2))= a (sec theta xx sec^(2) theta+ tan theta xx sec theta ten theta)`
` -b[cosec theta (-cosec^(2) theta)+cot theta(-cosec theta cot theta)]`
` = a sec theta(sec^(2) theta + tan ^(2) theta)+b cosec theta (cosec^(2) theta+ cot ^(2) theta)`
since` 0 lt theta ltpi/2` so the trigonometric ratios will be positive and ` a gt 0, b gt 0 `
`:. (d^(2)l)/(d theta^(2)) gt 0`
` rArr" when" tan theta = (b/a)^(1/3)`

then l is minimum,
` :.` minimum value of
` l = a sec theta + b cosec theta`
`asqrt(a^(2/3)+b^(2/3))/a^(1/3)+bsqrt(a^(2/3)+b^(2/3))/b^(1/3)`
` =sqrt(a^(2/3)+b^(2/3))(a^(2/3)+b^(2/3))=(a^(2/3)+b^(2/3))^(3/2)`.