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Application of Derivatives

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Long Answer Type

351.

Find the largest possible area of a right-angled triangle whose hypotenuse is 5 cm long.

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352. 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 $open parentheses straight a to the power of 2 over 3 end exponent plus straight b to the power of 2 over 3 end exponent close parentheses to the power of 3 over 2 end exponent.$

Let ABC be given triangle in which $Syntax error from line 1 column 49 to line 1 column 152. Unexpected '<mlongdiv '.$ , so that AC = l is hypotenuse.

Let P be any point on AC and $PM perpendicular BC comma space space PN perpendicular AB space so space that space PM space equals space straight b comma space space space PN space equals space straight a.$
Let $Syntax error from line 1 column 49 to line 1 column 152. Unexpected '<mlongdiv '.$

Now, $l space equals space AC space equals space AP plus PC$
$equals space straight a space secθ space plus space space straight b space cosecθ comma space space where space 0 less than straight theta less than straight pi over 2 therefore space space space space space dl over dθ space equals space straight a space secθ space tanθ minus space space straight b space cosecθ space cotθ space space space For space maximum space or space minimum comma space space space space space space space space dl over dθ space equals space 0$

$rightwards double arrow space space space straight a space secθ space tanθ space minus space straight b space cosecθ space cotθ space equals space 0 rightwards double arrow space space space straight a space secθ space tanθ space equals space straight b space cosecθ space cot space straight theta rightwards double arrow space space space space straight a 1 over cosθ. space fraction numerator sin space straight theta over denominator cos space straight theta end fraction space equals space straight b. space fraction numerator 1 over denominator sin space straight theta end fraction. space fraction numerator cos space straight theta over denominator sin space space straight theta end fraction rightwards double arrow space space space fraction numerator sin cubed space straight theta over denominator cos cubed space straight theta end fraction space equals space straight b over straight a space space rightwards double arrow space space space tan cubed straight theta space equals space straight b over straight a space space space rightwards double arrow space space space space tan space straight theta space equals space open parentheses straight b over straight a close parentheses to the power of 1 third end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$

Now, $fraction numerator straight d squared straight l over denominator dθ squared end fraction space equals space space straight a space sec space straight theta space. sec squared straight theta space plus space straight a space secθ space tan squared straight theta space plus space straight b space cosecθ space cot squared straight theta space plus space straight b space cosec cubed straight theta$
$space equals space straight a space secθ space left parenthesis sec squared straight theta space plus space tan squared straight theta right parenthesis space plus space straight b space cosecθ thin space left parenthesis cosec squared straight theta space plus space cot squared straight theta right parenthesis thin space greater than space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space space 0 less than 0 less than straight pi over 2 comma space space therefore space space secθ comma space tanθ space are space positive comma space Also comma space straight a comma space straight b space are space positive. close square brackets therefore space space space space space space space straight l space is space minimum space when space tanθ space equals space open parentheses straight b over straight a close parentheses to the power of 1 third end exponent therefore space space space space space space minimum space value space of$

$straight l space equals space straight a fraction numerator square root of straight a to the power of begin display style 2 over 3 end style end exponent plus straight b to the power of begin display style 2 over 3 end style end exponent end root over denominator straight a to the power of begin display style 1 third end style end exponent end fraction plus straight b fraction numerator square root of straight a to the power of begin display style 2 over 3 end style end exponent plus straight b to the power of begin display style 2 over 3 end style end exponent end root over denominator straight b to the power of begin display style 1 third end style end exponent end fraction$

$equals space straight a to the power of 2 over 3 end exponent square root of straight a to the power of 2 over 3 end exponent plus straight b to the power of 2 over 3 end exponent end root space plus space straight b to the power of 2 over 3 end exponent square root of straight a to the power of 2 over 3 end exponent plus straight b to the power of 1 third end exponent end root space equals space square root of straight a to the power of 2 over 3 end exponent plus straight b to the power of 2 over 3 end exponent end root space open parentheses straight a to the power of 2 over 3 end exponent plus straight b to the power of 2 over 3 end exponent close parentheses equals space open parentheses straight a to the power of 2 over 3 end exponent plus straight b to the power of 2 over 3 end exponent close parentheses to the power of 3 over 2 end exponent$

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353. Find the maximum-area of an isosceles triangle inscribed in the ellipse

straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1

with its vertex at one end of the major axis.

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Short Answer Type

354.

Let AP and BQ be two vertical poles at points A and B, respectively. If AP = 16 m. BQ = 22 m and AB = 20 m, then find the distance of a point R on AB from the point A such that RP² + RQ² is minimum.

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