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Problem Of The Week 45

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Find the largest value for for pairs of real numbers which satisfy


Written by Implex

October 5, 2008 at 10:53 am

10 Responses

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  1. (x-3)^2 +(y-3)^2 = 6

    let y = kx
    (x-3)^2 + (kx-3)^2 = 6
    x^2+9-6x + k^2x^2 +9 -6kx = 6
    (1+k^2)x^2 -(6+6k)x +12 = 0

    Since x is real


    (6+6k)^2 – 48(1+k^2) >=0

    Max value of k will be between 5 and 6


    Is this the correct approach?


    October 5, 2008 at 2:44 pm

  2. no it is not good approach


    October 5, 2008 at 4:20 pm

  3. the equation mentioned is of a circle with center at 3,3 and radius sqrt(6)
    so y max= 3+sqrt(6) x min= 3-sqrt(6)

    y/x= 5+2sqrt(6)

    is this approach, fine???


    October 5, 2008 at 6:33 pm

  4. This is surely much better approach ankaj – But i think x and y should be such that they satisfy the circle’s equation together. So, for y= 3+sqrt6 x=3

    If we divide the circle into 4 parts(shift axis to 3,3) then we can see that the maximum value of y/x can be in quad 1 or quad 2(by observing we can eliminate quad 3 and 4, as y/x increases more rapidly in quad 1 as compared to quad 4, and in quad 3 y decreases while x increases) .

    Just by little more observation we find that y/x will be maximum at y=3, x=3-sqrt6 (left most point on the circle). Good approach i must say 🙂

    Hence, y/x= 3 + sqrt6.


    October 5, 2008 at 7:50 pm

  5. y/x =3+root6 sint/3+root6cost

    differenciatin i got
    sint +cost = 2/root6

    y/x = 7+root5 /7-root5


    October 6, 2008 at 2:45 pm

  6. we can think of it sin x ->1
    cos x->0

    3+root6 /3 = approx2.. something less than 2


    October 6, 2008 at 2:45 pm

  7. just browsing thru this site today…. implex bhai……sahi qns hai…
    ne ways this is a nice qn n can be solved completely by graphical method… no need of using trigo/calculus… a simple qn actually..
    draw a graph of the circle with the given eq.
    now we want to maximise y/x….. let y=kx
    now we choose a ‘k’ such that the graph of y=kx touches the circle.
    now the max such ‘k’ can be found if the said graph (y=kx) is tangent to the circle. it is tangent at 2pts one of which is max & the other minimum.
    if iv not made a calc mistake: the max y/x comes to (3+sqrt2)…
    nice qns..will visit this blog more frequently now ons..


    October 14, 2008 at 6:49 am

  8. yupp but using coordinate, we alwasy know what are coods of points on a circle, we are done without any knew work


    October 14, 2008 at 9:44 am

  9. We can see that the mentioned equation represents a circle. Hence the maximum value of y/x is the greater of the slopes of the two common tangents from origin to the circle.

    Hence max(y/x) = tan(45+arcsin(sqrt(6)/3sqrt(2))

    tan(arcsin(1/sqrt(3)) = 1/sqrt(2)

    Therefore, max(y/x) = 3+2sqrt(2)


    October 16, 2008 at 7:09 am

  10. vijay is the best……..!!!


    October 29, 2008 at 2:59 pm

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