Let a and b be positive real numbers. If x^2 + y^2<=1 then what is the largest that ax+by can get?

By the Cauchy-Schwartz inequality, we have (x2 + y2)(a2+b2) >= (ax+by)2.
This can be transformed into (ax+by)2 <= (x2 + y2)(a2+b2) <= 1 * (a2+b2) <= (a2+b2). Hence ax + by <= sqrt(a^2 + b^2) and the equality is achieved when there exists ay = bx.

TD
Answered by Tutor135762 D. MAT tutor

4364 Views

See similar MAT University tutors

Related MAT University answers

All answers ▸

How many distinct real roots does the equation x^3 − 30x^2 + 108x − 104 = 0 have?


Let r and s be integers. Then ( 6^(r+s) x 12^(r-s) ) / ( 8^(r) x 9^(r+2s) ) is an integer when: (a) r+s <= 0, (b) s <= 0, (c) r <= 0, (d) r >= s.


Let a and b be positive integers such that a+b = 20. What is the maximum value that (a^2)b can take?


(Note this is the kind of exercise I would ask someone who is doing further maths and especially someone MAT/STEP) Sketch the graph of y=sin(1/x)


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning