Application of Derivatives

Question

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

Solution

Let ABC be right-angled triangle in which ∠ABC = 90°, AB = x, AC = y (constant).
$angle ABC space equals space 90 degree comma space space AB space equals straight x comma space space AC space equals space straight y space space left parenthesis constant right parenthesis. therefore space space space space BC space equals space square root of straight y squared minus straight x squared end root$
Let S denote the perimeter of the triangle.
$therefore space space space straight S space equals space straight x plus straight y plus square root of straight y squared minus straight x squared end root space space space dS over dx equals 1 plus 0 plus fraction numerator negative 2 straight x over denominator 2 square root of straight y squared minus straight x squared end root end fraction space equals space 1 minus fraction numerator straight x over denominator square root of straight y squared minus straight x squared end root end fraction For space straight S space to space be space maximum space or space minimum comma space dS over dx space equals space 0 therefore space space space space space space space 1 minus fraction numerator straight x over denominator square root of straight y squared minus straight x squared end root end fraction space equals space 0 comma space space space space space or space space square root of straight y squared minus straight x squared end root space equals space straight x therefore space space space space space space space space straight y squared minus straight x squared space equals space straight x squared space space space space space space space or space space space 2 straight x squared space equals space straight y squared space space space space space space space space rightwards double arrow space space space space space straight x squared space equals space straight y squared over 2 therefore space space space space space space space straight x space equals space fraction numerator straight y over denominator square root of 2 end fraction space as space straight x space is space plus ve$
$fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space 0 minus fraction numerator square root of straight y squared minus straight x squared end root. space space 1 space space minus straight x 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 left parenthesis square root of straight y squared minus straight x squared end root right parenthesis squared end fraction space equals space minus fraction numerator begin display style fraction numerator 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 squared 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 space space space space space space space space space space space equals negative fraction numerator straight y squared minus straight x squared plus straight x squared over denominator left parenthesis straight y squared minus straight x squared right parenthesis end fraction space equals space minus fraction numerator straight y 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 When space straight x space equals fraction numerator straight y over denominator square root of 2 end fraction. space fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space minus straight y squared over 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 space equals space fraction numerator straight y squared over denominator begin display style fraction numerator straight y cubed over denominator 2 square root of 2 end fraction end style end fraction space equals space minus fraction numerator 2 square root of 2 over denominator straight y end fraction less than 0$
$therefore space space space space space straight S space is space maximum space when space straight x space equals space fraction numerator straight y over denominator square root of 2 end fraction therefore space space space space AB space equals space fraction numerator straight y over denominator square root of 2 end fraction space space space space space space space space space BC space equals space square root of straight y squared minus straight x squared end root space equals space square root of straight y squared minus straight y squared over 2 end root space equals space fraction numerator straight y over denominator square root of 2 end fraction. therefore space space space space space AB space equals space BC therefore space space space space space space increment ABC space is space isosceles$

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Application Of Derivatives

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

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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