## Problem Of The Week 45

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

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For All Your Quant Queries

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

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Written by Implex

October 5, 2008 at 10:53 am

Posted in Geometry, Problem of the week

Tagged with Geometry, Problem of the week

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(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?

AmitOctober 5, 2008 at 2:44 pm

no it is not good approach

outtimedOctober 5, 2008 at 4:20 pm

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???

ankajOctober 5, 2008 at 6:33 pm

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.

AmitOctober 5, 2008 at 7:50 pm

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

differenciatin i got

sint +cost = 2/root6

y/x = 7+root5 /7-root5

VarunOctober 6, 2008 at 2:45 pm

we can think of it sin x ->1

cos x->0

so

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

VarunOctober 6, 2008 at 2:45 pm

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

nbangalorekarOctober 14, 2008 at 6:49 am

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

outtimedOctober 14, 2008 at 9:44 am

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)

VijayOctober 16, 2008 at 7:09 am

vijay is the best……..!!!

wolverineOctober 29, 2008 at 2:59 pm