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

3634 Views

See similar MAT University tutors

Related MAT University answers

All answers ▸

Circle the correct letter: The equation x^3 - 30x^2 + 108x - 104 = 0 has a) No real roots; b) Exactly one real root; c) Three distinct real roots; d) A repeated root.


We define the digit sum of a non-negative integer to be the sum of its digits. For example, the digit sum of 123 is 1 + 2 + 3 = 6. Let n be a positive integer with n < 10. How many positive integers less than 1000 have digit sum equal to n?


A trillion is 10^12. Which of the following is bigger: the three trillionth root of 3 or the two trillionth root of 2? You may assume that if 0 < x < y, then 0 < x^n < y^n for integer values of n greater than or equal to 1.


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.


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:

© 2025 by IXL Learning