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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.

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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.

Let R be a point on AB such that AR = x metres, RB = (20 – x) metres
∴ RP² = AR² + AP²= x² + (16)²= x²+ 256
and RQ² = RB² + BQ²= (20 – x)² + (22)²= 400+ x² – 40 x + 484
= x²– 40 x + 884
Let y = RP² + RQ^{2

$therefore space space space space space straight y space equals space left parenthesis straight x squared plus 256 right parenthesis space plus left parenthesis straight x squared minus 40 straight x plus 884 right parenthesis therefore space space space space space straight y space equals space 2 straight x squared minus 40 straight x plus 1140 space space space space space space space space space space space space dy over dx space equals space 4 straight x minus 40 space space space space space space space space space space space space space dy over dx space equals space 0 space space space space rightwards double arrow space space space space 4 straight x minus 40 space equals space 0 space space space space rightwards double arrow space space space space straight x space equals space 10 space space space space space space space space space space space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space 4 When space straight x space equals space 10 comma space space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space 4 greater than 0 therefore space space space space space space straight y space is space minimum space when space straight x space space equals 10 therefore space space space space space space straight y space is space minimum space when space AR space equals space 10 space metres.$}