IIT JEE , AIEEE Forum - Mathematics , Trignometry

cool_naman92

a²+b²=4 and x²+y²=9

then find maximum and minimum value of ax+by

rockey

a^2+b^2)(x^2+y^2) =a^2 x^2 +b^2 y^2+a^2y^2 + b^2x^2 =36
(ax+by)^2=36_(ay-bx)^2
i hope u can solve it

rate me if useful

hsbhatt

Just use .

a/2 = sinp; b/2 = cosp; x/3 = sinq; y/3 = cosq

and you will get (ax+by)/6 = cos(p-q)

-1

cos(p-q)

1

Hence -6

ax+by

6

puneet

hii

yes HS bhatt is perfectly correct .. this hsud strike u as soon as u see hte question ..

the equation of circle clearly suggets that put it in polar co-ordinates ..

cheers

hsbhatt

Actually, if you are more conversant with inequalities, this is a direct application of the Cauchy-Schwarz Inequality.

(a²+b²) (x²+y²)

(ax+by)²

Hence (ax+by)²

36

or -6

ax+by

6

karthik2789

max value =6
min value= -6

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