Question

If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum when the angle between them is $60 degree.$

Solution

Let ABC be the right angle triangle with base b and hypotenuse h.
Given that b+ h = k
Let A be the area of the right triangle.
$straight A equals 1 half cross times straight b cross times square root of straight h squared minus straight b squared end root rightwards double arrow space space straight A squared space equals space 1 fourth straight b squared left parenthesis straight h squared minus straight b squared right parenthesis rightwards double arrow space straight A squared equals space space straight b squared over 4 open parentheses open parentheses straight k minus straight b close parentheses squared minus straight b squared close parentheses space space space space space space open square brackets because space straight h space equals space straight k minus straight b close square brackets rightwards double arrow straight A squared space equals space straight b squared over 4 open parentheses straight k squared plus straight b squared minus 2 kb minus straight b squared close parentheses rightwards double arrow straight A squared space equals space straight b squared over 4 left parenthesis straight k squared minus 2 kb right parenthesis rightwards double arrow straight A squared equals space fraction numerator straight b squared straight k squared minus 2 kb cubed over denominator 4 end fraction$
Differentiating the above function with respect to be, we have
$2 straight A dA over db equals fraction numerator 2 bk squared minus 6 kb squared over denominator 4 end fraction space... left parenthesis 1 right parenthesis rightwards double arrow space space dA over db equals fraction numerator bk squared minus 3 kb squared over denominator 2 straight A end fraction$
For the area to be maximum, we have
$dA over db equals 0$
$rightwards double arrow space bk squared minus 3 kb squared space equals 0 rightwards double arrow space bk space equals space 3 straight b squared rightwards double arrow space straight b space equals space straight k over 3$
Again differentiating the function in equation (1), with respect to b, we have
$2 open parentheses dA over db close parentheses squared plus 2 straight A fraction numerator straight d squared straight A over denominator db squared end fraction equals fraction numerator 2 straight k squared minus 12 kb over denominator 4 end fraction space... left parenthesis 2 right parenthesis$
Now substituting $dA over db equals 0 space and space straight b space equals space straight k over 3$ in equation (2), we have
$2 straight A fraction numerator straight d squared straight A over denominator db squared end fraction space equals space fraction numerator 2 straight k squared minus 12 straight k open parentheses begin display style straight k over 3 end style close parentheses over denominator 4 end fraction space space space rightwards double arrow 2 straight A fraction numerator straight d squared straight A over denominator db squared end fraction space equals space fraction numerator 6 straight k squared minus 12 straight k squared over denominator 12 end fraction space space rightwards double arrow space 2 straight a fraction numerator straight d squared straight A over denominator db squared end fraction equals negative straight k squared over 2 space space space rightwards double arrow space fraction numerator straight d squared straight A over denominator db squared end fraction equals negative fraction numerator straight k squared over denominator 4 straight A end fraction less than 0$
Thus area is maximum at $straight b equals straight k over 3.$
Now, $straight h equals straight k minus straight k over 3 equals fraction numerator 2 straight k over denominator 3 end fraction$
Let $straight theta$ be the angle between the base of the triangle and the hypotenuse of the right angle.
$Thus comma space cosθ equals straight b over straight h equals fraction numerator begin display style straight k over 3 end style over denominator begin display style fraction numerator 2 straight k over denominator 3 end fraction end style end fraction equals 1 half rightwards double arrow space straight theta equals space cos to the power of negative 1 end exponent open parentheses 1 half close parentheses equals space straight pi over 3$

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Application Of Derivatives

If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum when the angle between them is $60 degree.$

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