How do you maximize and minimize #f(x,y)=x^2-y/x# constrained to #0<=x+y<=1#?

1 Answer
Feb 16, 2016

There is no upper bound for #f#.

The lower bound is #f(0,0) = 1#

Explanation:

Substitute #x + y = u#.

Now, optimize

#g(x,u) = x^2 - (u-x)/x#

#= x^2 + 1 - u/x#

With the only condition being #0 <= u <= 1#.

When #x# is large, the #x^2# term dominates and the #u/x# term becomes insignificant. By increasing #x#, #f# can be made arbitrarily large.

#frac{del g}{del x} = 2x + u/x^2#

From the first equation, we see that for a given value of #u#, minimum occurs when #frac{del g}{del x} = 0#. Therefore,

#x = -root(3){u/2}#

So to find the minimum of #g#, we replace all instances of #x# with #-root(3){u/2}#.

#g(u) = (-root(3){u/2})^2 + 1 - u/(-root(3){u/2})#

# = 2^{-2/3}*u^{2/3} + 1 + root(3)2*u^{2/3}#

# = 3root(3){u^2/4} + 1 #

Even without differentiating, it is quite easy to see that #g# is increasing for #u in [0,1]#

The mimimum corresponds to #u = 0#, and consequently, #x = 0#, and #y = 0#.

The absolute minimum is #f(0,0) = 1#