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Finding extreme points subject to constraint

Hello,
I need to find (if there are) minimum and maximum values of the following function: z=1x+1yz=\frac{1}{x}+\frac{1}{y}
subject to constraint: 1x2+1y2=1a2 \frac{1}{{x}^{2}}+\frac{1}{{y}^{2}}=\frac{1}{{a}^{2}} (a≠0 a\neq 0 )

I think there are no extrema, but I do not know how to show it.
(edited 11 years ago)
Reply 1
Avatar for nuodai
nuodai
17
Based on what you've said, z=1a2z=\dfrac{1}{a^2} so it is constant, so there are no extrema. Are you sure you copied out the question correctly?
Reply 2
Avatar for msokol
msokol
OP
Oops, sorry, changed it
if (x,y) gives the maximum then (-x,-y) will give the minimum. So let's find the maximum. The maximum of z will be the maximum of z^2, and z^2 = 1/x^2 + 1/y^2 + 2/xy = 1/a^2 + 2/xy. So it reduces to finding the maximum of 2/xy subject to 1/x^2 + 1/y^2 = 1/a^2, which can be done by applying the AM-GM inequality to {1/x^2,1/y^2}.

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