anonymous
  • anonymous
Find the maximum value of x^2 + y^2 which satisfy the equation x^2 + y^2 - 10x - 4y - 20 = 0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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freckles
  • freckles
have you tried lagrange multiplers
ganeshie8
  • ganeshie8
https://www.desmos.com/calculator/3j16xhdqpa
anonymous
  • anonymous
never study with lagrange yet :( can we solve with other method ?

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anonymous
  • anonymous
@ganeshie8. i cant understand what the pic means ?
anonymous
  • anonymous
here the choices answer : a. 78 + 14 sqrt(29) b. 76 + 14 sqrt(29) c. 76 + 14 sqrt(26) d. 74 + 14 sqrt(29) e. 72 + 14 sqrt(26) which one the correct option and also the steps to get it
baru
  • baru
maybe we should factor the second equation into an equation of circle with shifted origin, solve for y, and substitute in the func. which needs maximizing
triciaal
  • triciaal
|dw:1446359811486:dw|
anonymous
  • anonymous
x^2 + y^2 = k^2 ----> x^2 = k^2 - y^2 x^2 + y^2 - 10x - 4y - 20 = 0 k^2 - y^2 + y^2 - 10x - 4y - 20 = 0 k^2 - 10x - 4y - 20 = 0 how to get max value of k ?
baru
  • baru
seems like the graph interpretation is the only way to go. max value of f(x)= square of ( point on the given circle farthest from the origin) point on the given circle farthest from the origin= (sqrt(5^2 + 2^2)+ radius)
ganeshie8
  • ganeshie8
\((x-5)^2 + (y-2)^2 = 49\) Consider the vectors \((10, 4)\) and \((x-5, y-2)\). Appealing to cauchy schwarz inequality : \(10(x-5) + 4(y-2) \le \sqrt{10^2+4^2}\sqrt{(x-5)^2+(y-2)^2}\) see if you can manage the rest...
freckles
  • freckles
hmmm... does anyone know how to show: \[\text{ the max of } (x-5) \text{ is } \frac{ 7 \cdot 5}{\sqrt{29}} \\ \text{ while the max of } (y-2) \text{ is } \frac{7 \cdot 2}{\sqrt{29}}\] just asking these numbers are interesting sqrt(49) is 7 and 29 equal equal to 5^2+2^2
imqwerty
  • imqwerty
another method is to convert the equation of circle into parametric form to get the answer.
anonymous
  • anonymous
i get this question from the junior school problem math olympiad. please dont use hardest method :)
freckles
  • freckles
oh other than lagrange multiplers does anyone know how to show what I was asking...
ganeshie8
  • ganeshie8
then you are allowed to use cauchy-schwarz inequality right?
tkhunny
  • tkhunny
@baru gave it in its simplest solution. Simply observe that x^2+ y^2 is the squared distance formula (from the origin). \(7 + \sqrt{29}\) is the easily-observed maximum distance from the origin. Square that and you are done.
baru
  • baru
it gives 78+7sqrt29
freckles
  • freckles
I'm getting 78+14sqrt(29)
baru
  • baru
sorry, me too, 14sqrt29
triciaal
  • triciaal
|dw:1446361543350:dw|
baru
  • baru
if my logic isnt convincing, consider this the vector from origin to centre(C) and radius vector R will have max length when C and R both have same direction, if they are in the same direction, magnitute of (C+R)= (sqrt(5^2 + 4^2)+ radius)
triciaal
  • triciaal
seems like we all pick A
anonymous
  • anonymous
OK... thank you guys. i think i got it now
imqwerty
  • imqwerty
(x-5)^2 +(y-2)^2 =(7)^2 ^eq we have is the eq of circle any point x,y lying on circle can be given like this-\[x=h+r \cos(\theta)\]\[y=k+r \sin(\theta)\] origin cordinates->(h,k) radius->r \[x^2+y^2=h^2+k^2+r^2+2r(hcos(\theta)+ksin(\theta))\] \[x^2+y^2=(-5)^2+(-2)^2+7^2+2(-5\cos(\theta) -2\sin(\theta))\] \[x^2+y^2=78+2\sqrt{29}\]
imqwerty
  • imqwerty
we got the center cordinates and radius by comparing our equation-(x-5)^2 +(y-2)^2 =(7)^2 with this standard equation of the circle- \[(x-h)^2+(y-k)^2=r^2\] h,k->center cordinates r->radius
baru
  • baru
you have parametrized the distance function to the circle?
baru
  • baru
*distance squared
imqwerty
  • imqwerty
yes i used the parametric form to denote any point x,y lying on the circle |dw:1446359372917:dw|
baru
  • baru
got it :)
imqwerty
  • imqwerty
(:

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