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Question Number 135925 by liberty last updated on 17/Mar/21
minimum value of (√((x+6)^2 +25)) +(√((x−6)^2 +121))  equal to?
minimumvalueof(x+6)2+25+(x6)2+121equalto?
Answered by john_santu last updated on 17/Mar/21
 Minimum value of   (√((x+6)^2 +25)) + (√((x−6)^2 +121)) equal to  standart method  f(x)=(√((x+6)^2 +25)) + (√((x−6)^2 +121))  f ′(x)= ((x+6)/( (√((x+6)^2 +25)))) + ((x−6)/( (√((x−6)^2 +121)))) = 0   ((x+6)/( (√((x+6)^2 +25)))) = ((6−x)/( (√((x−6)^2 +121))))  ⇒ (((x+6)^2 )/((x+6)^2 +25)) = (((6−x)^2 )/((6−x)^2 +121))  ⇒(((x+6)^2 +25)/((x+6)^2 )) = (((6−x)^2 +121)/((6−x)^2 ))  ⇒ ((5/(x+6)))^2 = (((11)/(6−x)))^2   we get  { (((5/(x+6)) = ((11)/(6−x)))),(((5/(x+6)) = ((11)/(x−6)))) :}  (1) 30−5x = 11x+66 , 16x=−36,x=−(9/4)  (2)5x−30 = 11x+66, 6x=−96,x=−16  for x=−(9/4)⇒f_1 =(√((((15)/4))^2 +25)) +(√((−((33)/4))^2 +121))   = 20 (minimum value )  for x=−16⇒f_2 =(√((−10)^2 +25))+(√((−22)^2 +121))  = 16(√5) ≈ 35.77 (maximum value)
Minimumvalueof(x+6)2+25+(x6)2+121equaltostandartmethodf(x)=(x+6)2+25+(x6)2+121f(x)=x+6(x+6)2+25+x6(x6)2+121=0x+6(x+6)2+25=6x(x6)2+121(x+6)2(x+6)2+25=(6x)2(6x)2+121(x+6)2+25(x+6)2=(6x)2+121(6x)2(5x+6)2=(116x)2weget{5x+6=116x5x+6=11x6(1)305x=11x+66,16x=36,x=94(2)5x30=11x+66,6x=96,x=16forx=94f1=(154)2+25+(334)2+121=20(minimumvalue)forx=16f2=(10)2+25+(22)2+121=16535.77(maximumvalue)
Commented by mr W last updated on 17/Mar/21
maximum doesn′t exist, since  lim_(x→−∞) f(x)=+∞  lim_(x→+∞) f(x)=+∞
maximumdoesntexist,sincelimxf(x)=+limx+f(x)=+
Answered by mr W last updated on 17/Mar/21
Commented by mr W last updated on 17/Mar/21
we can solve this problem without  using calculus.  OA=(√((x+6)^2 +5^2 ))  OB=(√((x−6)^2 +11^2 ))  mininum from OA+OB is the  straight line AB=(√((6+6)^2 +(5+11)^2 ))  =20, when the origin is at point C, or  x=−(9/4).
wecansolvethisproblemwithoutusingcalculus.OA=(x+6)2+52OB=(x6)2+112mininumfromOA+OBisthestraightlineAB=(6+6)2+(5+11)2=20,whentheoriginisatpointC,orx=94.
Commented by mr W last updated on 17/Mar/21
generally using this we get  mininum from  (√((x+a)^2 +c^2 ))+(√((x+b)^2 +d^2 )) is  (√((a−b)^2 +(c+d)^2 ))  (assume c,d ≥0)
generallyusingthiswegetmininumfrom(x+a)2+c2+(x+b)2+d2is(ab)2+(c+d)2(assumec,d0)
Commented by liberty last updated on 17/Mar/21
waw...it follows that if  (√((x+3)^2 +36)) + (√((x−3)^2 +49))  the method valid sir?
wawitfollowsthatif(x+3)2+36+(x3)2+49themethodvalidsir?
Commented by mr W last updated on 17/Mar/21
yes.  an other example:  [(√((x+3)^2 +10)) + (√((x+8)^2 +20))]_(min)   =(√((8−3)^2 +((√(10))+(√(20)))^2 ))=(√(55+20(√2)))  you can check it with Grapher or  using calculus.
yes.anotherexample:[(x+3)2+10+(x+8)2+20]min=(83)2+(10+20)2=55+202youcancheckitwithGrapherorusingcalculus.
Commented by liberty last updated on 17/Mar/21
great....
great.
Commented by Dwaipayan Shikari last updated on 17/Mar/21
Newton used to have Geometrical Method of Calculus
NewtonusedtohaveGeometricalMethodofCalculus

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