Application of Derivatives

Question

Prove that the area of a right angled triangle of given hypotenuse is maximum when the triangle is isosceles.

Solution

Let ABC be right angled triangle in which
$angle ABC space equals space 90 degree$
$AB space equals space straight x comma space space space space AC space equals space straight y$
$therefore space space space space space space space space BC space equals space square root of straight y squared minus straight x squared end root space where space straight y space is space hypotenuse.$

Let ∆ denote the area of triangle ABC.
$therefore space space space space space increment space equals space 1 half BC. space AB space equals space 1 half square root of straight y squared minus straight x squared end root space space. straight x space space space space fraction numerator straight d increment over denominator dx end fraction space equals space 1 half open square brackets straight x. space straight d over dx open parentheses square root of straight y squared minus straight x squared end root close parentheses space plus space square root of straight y squared minus straight x squared end root. space straight d over dx left parenthesis straight x right parenthesis close square brackets space space space space space space space space space space space space space space space equals space 1 half open square brackets straight x. space fraction numerator negative 2 straight x over denominator 2 square root of straight y squared minus straight x squared end root end fraction plus square root of straight y squared minus straight x squared end root. space 1 close square brackets space space space space space space space left square bracket Here space straight y space is space constant right square bracket space space space space space space space space space space space space space space space equals space 1 half open square brackets fraction numerator negative straight x squared plus straight y squared minus straight x squared over denominator square root of straight y squared minus straight x squared end root end fraction close square brackets space equals space 1 half open square brackets fraction numerator left parenthesis straight y squared minus 2 space straight x squared over denominator square root of straight y squared minus straight x squared end root end fraction close square brackets Now space space space fraction numerator straight d increment over denominator dx end fraction space equals space 0 space space gives space us space space space space space 1 half open square brackets fraction numerator straight y squared minus 2 straight x squared over denominator square root of straight y squared minus straight x squared end root end fraction close square brackets space space equals 0 comma space space space space space space or space straight y squared space minus space 2 straight x squared space equals space equals 0 therefore space space space space space straight x squared space equals space space straight y squared over 2 space space space space space space space space space space space space space space rightwards double arrow space space space space space space straight x space equals plus-or-minus fraction numerator straight y over denominator square root of 2 end fraction.$
Rejecting $straight x space equals space minus fraction numerator straight y over denominator square root of 2 end fraction comma space space space we space get comma space space space space straight x space equals space fraction numerator straight y over denominator square root of 2 end fraction.$
$fraction numerator straight d squared increment over denominator dx squared end fraction space equals space 1 half open square brackets fraction numerator square root of straight y squared minus straight x squared end root. space left parenthesis negative 4 straight x right parenthesis space minus space left parenthesis straight y squared minus 2 straight x squared right parenthesis. space space begin display style fraction numerator negative 2 straight x over denominator 2 square root of straight y squared minus straight x squared end root end fraction end style over denominator straight y squared minus straight x squared end fraction close square brackets space space space space space space space space space space space space space space equals space fraction numerator 1 over denominator space 2 end fraction open square brackets fraction numerator begin display style fraction numerator negative 4 straight x square root of straight y squared minus straight x squared end root over denominator 1 end fraction end style plus begin display style fraction numerator straight x left parenthesis straight y squared minus 2 straight x squared right parenthesis over denominator square root of straight y squared minus straight x squared end root end fraction end style over denominator straight y squared minus straight x squared end fraction close square brackets space space space space space space space space space space space space space space space space equals space 1 half open square brackets fraction numerator negative 4 straight x left parenthesis straight y squared minus straight x squared right parenthesis space plus space straight x left parenthesis straight y squared minus 2 straight x squared right parenthesis over denominator left parenthesis straight y squared minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets space space space space space space space space space space space space space space space$
$equals space 1 half open square brackets fraction numerator negative 4 xy squared plus 4 straight x cubed plus xy squared minus 2 straight x cubed over denominator left parenthesis straight y squared minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets space equals space open square brackets fraction numerator 2 straight x cubed minus 3 xy squared over denominator left parenthesis straight y squared minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets$
$When space straight x space equals space fraction numerator straight y over denominator square root of 2 end fraction comma space space space fraction numerator straight d squared increment over denominator dx squared end fraction space equals space 1 half open square brackets fraction numerator 2. space begin display style fraction numerator straight y cubed over denominator 2 square root of 2 end fraction end style space minus 3 space straight y squared. begin display style fraction numerator straight y over denominator square root of 2 end fraction end style over denominator open parentheses straight y squared minus begin display style straight y squared over 2 end style close parentheses to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets space space space space space space space space space space space space equals 1 half open square brackets fraction numerator begin display style fraction numerator straight y cubed over denominator square root of 2 end fraction end style minus begin display style fraction numerator 3 straight y cubed over denominator square root of 2 end fraction end style over denominator begin display style fraction numerator straight y cubed over denominator 2 square root of 2 end fraction end style end fraction close square brackets space equals space 1 half open square brackets fraction numerator negative square root of 2 space straight y cubed over denominator begin display style fraction numerator straight y cubed over denominator 2 square root of 2 end fraction end style end fraction close square brackets space equals space minus 2 space less than space 0$
$therefore space space space space increment space is space maximum space when space straight x space equals space fraction numerator straight y over denominator square root of 2 end fraction therefore space space space space AB space equals space fraction numerator straight y over denominator square root of 2 end fraction space space space space space space space space space space BC space equals space square root of straight y squared minus straight x squared end root space equals space square root of straight y squared minus straight y squared over 2 end root space equals space fraction numerator straight y over denominator square root of 2 end fraction therefore space space space space AB space equals space BC space space space space space rightwards double arrow space space space space increment ABC space is space isosceles.$
∴ area of a right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.

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Application Of Derivatives

Prove that the area of a right angled triangle of given hypotenuse is maximum when the triangle is isosceles.

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