Turn on thread page Beta

Partial differentiation help!!! watch

Announcements
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hi

    Stuck on this one question, would really appreciate some help.

    Show that the minimum value of:

     c^3 (\frac{1}{x} + \frac{1}{y}) + xy

    is

     3c^2

    where c is a constant

    Calling the above equation z i tried partially differentiating it in terms of x and y, and then equalling both equations to zero. Tried different combinations of canceling out and re-arranging but cant seem to get rid of the the x's and y's.

    Would appreciate any help,

    Thanks
    Offline

    18
    ReputationRep:
    Post your working. It should only be a few lines.
    Offline

    13
    ReputationRep:
    By\   symmetry

    x=y

    f(x)=\frac{2c^3}{x}+x^2

    f'(x)=\frac{-2c^3}{x^2}+2x

    At\   minima, f'(x)=0

    \frac{-2c^3}{x^2}+2x=0

    x^3=c^3

    x=c

    So,

    f(x)=\frac{2c^3}{c}+c^2

    f(x)=2c^2+c^2

    f(x)=3c^2
    Offline

    0
    ReputationRep:
    (Original post by thomaskurian89)
    By\   symmetry
    What does that exactly mean? And how do you recognise it?

    Thanks
    Offline

    13
    ReputationRep:
    The given function is symmetric in x and y, ie the function does not change if x and y are interchanged.
    Offline

    0
    ReputationRep:
    Thank Youu
    Offline

    13
    ReputationRep:
    Since we're on the topic,

    Is x = y assumption valid?
    • Thread Starter
    Offline

    0
    ReputationRep:
    yeh i got the same thing, i just made a simple mistake when dividing. Its quite an easy question infact.

    Thanks for all your replies.
    Offline

    18
    ReputationRep:
    Well, obviously it's true here. But there are lots of symmetric minimization problems with asymmetric solutions - for example

    Minimize xyz-(x+y+z) given that x^2+y^2+z^2 = 2

    has its minimum at x=y=1, z= 0 (plus 5 other obviously related solutions).

    So no, I don't see it's valid without justification.
    Offline

    13
    ReputationRep:
    (Original post by DFranklin)
    Well, obviously it's true here. But there are lots of symmetric minimization problems with asymmetric solutions - for example

    Minimize xyz-(x+y+z) given that x^2+y^2+z^2 = 2

    has its minimum at x=y=1, z= 0 (plus 5 other obviously related solutions).

    So no, I don't see it's valid without justification.
    So how then do you justify x = y in the the original problem?
    Offline

    18
    ReputationRep:
    The approach deepdawg originally suggested seems to work fine, as far as I can see.
    Offline

    13
    ReputationRep:
    (Original post by DFranklin)
    The approach deepdawg originally suggested seems to work fine, as far as I can see.
    Yes it did. Partial differentiation.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: August 27, 2008

University open days

  • University of East Anglia
    All Departments Open 13:00-17:00. Find out more about our diverse range of subject areas and career progression in the Arts & Humanities, Social Sciences, Medicine & Health Sciences, and the Sciences. Postgraduate
    Wed, 30 Jan '19
  • Aston University
    Postgraduate Open Day Postgraduate
    Wed, 30 Jan '19
  • Solent University
    Careers in maritime Undergraduate
    Sat, 2 Feb '19
Poll
Brexit: Given the chance now, would you vote leave or remain?
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

Equations

Best calculators for A level Maths

Tips on which model to get

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

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

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