At what point or points on the circle #x^2+y^2=1# does #f(x, y)=xy# have an absolute maximum, and what is that maximum?

2 Answers
Dec 20, 2017

See below.

Explanation:

Making a change of variables

#x=r cos theta#
#y=r sin theta# and substituting

#f(x,y) -> g(r, theta) = r^2 cos theta sin theta#

and the restriction gives #r = 1#

so the problem is reduced to

#min sin theta cos theta#

for #0 le theta < 2pi#

Now #(dg)/(d theta) = cos(2theta) = 0# gives the stationary points. or

#theta = pm pi/4 +k pi# and the maximum is for #theta = pi/4# and consequently

#x = cos(pi/4), y = sin(pi/4)#

Dec 21, 2017

Absolute max #= 1/2#

Explanation:

I wanted to provide an alternate way of answering this problem...

We know #x^2 + y^2 = 1 #

#=> y^2 = 1-x^2 #

#=> y = sqrt(1-x^2 ) #

So hence #f(x,y) = xsqrt(1-x^2 ) #

Now to find the maximum we must set the differential to 0:

#-------------------#

We must use our knowledge of the chain and product rule:

#if y = uv , y' = udv + vdu #

#if y = f(g(x)) , y' = f'(g(x)) * g'(x) #

#-------------------#

#f'(x,y) = sqrt(1-x^2) + x*( 1/2 * (1-x^2)^(-1/2) * -2x )#

#f'(x,y) = sqrt(1-x^2) - (x^2)/sqrt(1-x^2) #

Setting #f'(x,y) = 0 #:

#=> sqrt(1-x^2) - (x^2)/(sqrt(1-x^2)) = 0 #

#=> sqrt(1-x^2) = (x^2)/(sqrt(1-x^2) #

#=> 1-x^2 = x^2 #

#=> 1 = 2x^2 #

#=> 1/2 = x^2 #

#=> x = 1/sqrt2 #

Using #y = sqrt(1-x^2)#:

#=> y = 1/sqrt2 #

Max # xy => 1/sqrt2 * 1/sqrt2 = 1/2 #

We can check that this is a maximum by taking #f''(x,y) # and verifying that at #x= 1/sqrt2 # that #f''(1/sqrt2 ,y) < 0 #

As #(d^2y)/(dx^2) < 0 # means a maximum

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