Question

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

Answer 1

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`