Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    7
    ReputationRep:
    Let a and b be positive real numbers. And x^2 + y^2 is less than or equal to 1 . You have to find the maximum value of ax+by.

    This image was included as part of the solution. What I don't understand is why the diagonal length labelled is a/b.
    Attached Images
     
    Offline

    15
    ReputationRep:
    (Original post by stolenuniverse)
    Let a and b be positive real numbers. And x^2 + y^2 is less than or equal to 1 . You have to find the maximum value of ax+by.

    This image was included as part of the solution. What I don't understand is why the diagonal length labelled is a/b.
    The solution does seem rather opaque in its reasoning. Perhaps the easiest way to appreciate why that lenght is a/b is to appreciate it as the tangenet of certain angles.

    PS Especially as at a second glance it's clear the equation of the line has been incorrectly rearranged.
    Offline

    11
    ReputationRep:
    (Original post by stolenuniverse)
    Let a and b be positive real numbers. And x^2 + y^2 is less than or equal to 1 . You have to find the maximum value of ax+by.

    This image was included as part of the solution. What I don't understand is why the diagonal length labelled is a/b.
    The small triangle with side 1 is similar to the big triangle made with the line ax+by=c as the hypotenuse. The big triangle has vertical/horizontal sides ka and kb for some k (since the gradient is a/b), so the ratio of those sides is a/b which must also be the ratio of the corresponding sides in the smaller triangle.
    Offline

    11
    ReputationRep:
    (Original post by RichE)
    The solution does seem rather opaque in its reasoning. Perhaps the easiest way to appreciate why that lenght is a/b is to appreciate it as the tangenet of certain angles.

    PS Especially as at a second glance it's clear the equation of the line has been incorrectly rearranged.
    Has it? Those lengths look OK to me.
    Offline

    15
    ReputationRep:
    (Original post by atsruser)
    Has it? Those lengths look OK to me.
    The lengths are fine. Go to the full MAT solution to see the error of rearrangement.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.