Ultimate Quant Marathon Blog For IIM CAT

For All Your Quant Queries

Problem Of The Week 45

with 10 comments


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

Written by Implex

October 5, 2008 at 10:53 am

10 Responses

Subscribe to comments with RSS.

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

    let y = kx
    then,
    (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

    D>=0

    (6+6k)^2 – 48(1+k^2) >=0
    -12-12k^2+72k>=0
    -1-k^2+6k>=0
    1+k^2-6k<=0

    Max value of k will be between 5 and 6

    5<k<6

    Is this the correct approach?

    Amit

    October 5, 2008 at 2:44 pm

  2. no it is not good approach

    outtimed

    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???

    ankaj

    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.

    Amit

    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

    Varun

    October 6, 2008 at 2:45 pm

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

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

    Varun

    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..
    bye

    nbangalorekar

    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

    outtimed

    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)

    Vijay

    October 16, 2008 at 7:09 am

  10. vijay is the best……..!!!

    wolverine

    October 29, 2008 at 2:59 pm


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: