Range of $f(x,y)=\frac{4x^2+(y+2)^2}{x^2+y^2+1}$

Question

I am trying to find the range of this function:

$$f(x,y)=\frac{4x^2+(y+2)^2}{x^2+y^2+1}$$

So I think that means I have to find minima and maxima. Using partial derivatives gets messy, so I was wondering if I could do some change of variables to make it easier computationally. But no change of coordinates that I can think of have really simplified it much. If I set $2w=y+2$, then I get a problem below. Am I thinking of the right strategy, or is there something better I could do?

Answer 1

Idea: $$f(x,y) =\frac{4x^2+4y^2+4+(y+2)^2-4y^2-4}{x^2+y^2+1}$$

$$=4+\frac{-3y^2+4y}{x^2+y^2+1}$$

$$\leq 4+\frac{-3y^2+4y}{y^2+1}$$ if $-3y^2+4y\geq 0$ (else it is reversed),

$$= 1+\frac{4y+3}{y^2+1} =:g(y)$$

So you have to find a maximum value of $g$ on $[0,{4\over 3}]$.

Answer 2

It is $$f(x,y)\le 5$$ if $$4x^2+(y+2)^2\le 5x^2+5y^2+5$$ if $$0\le x^2+(2y-1)^2$$ and clearly we get $$f(x,y)\geq 0$$

Answer 3

See fig. below.

$5$ is the maximum. More precisely :

$$f(x,y):=\dfrac{4x^2+(y-2)^2}{x^2 + y^2 + 1}\leq 5 \tag{1}$$

with case of equality iff

$$(x,y)=(0,1/2)\tag{2}$$

Here is why : we form the "gap" between the LHS and RHS of (1) :

$$5-\dfrac{4x^2+(y-2)^2}{x^2 + y^2 + 1}=\dfrac{x^2+(2y - 1)^2}{x^2 + y^2 + 1}$$

This gap is always >0 with the exceptional case given by (2).

Besides, $0$ is clearly the minimum, realized for $(x,y)=(0,2)$.

Now the last question : is the range of values of function $f$ the whole interval $[0,5]$ ? Answer : yes by continuity.

Fig. 1 : Surface $z=f(x,y)$ with its unique minimum and maximum.

Answer 4

For $$f(x,y)=\frac{4x^2+(y+2)^2}{x^2+y^2+1}$$ partial derivatives are not so terrifying: \begin{align} D_1f(x,y)&=\frac{8x(x^2+y^2+1)-2x(4x^2+(y+2)^2)}{(x^2+y^2+1)^2}\\[4px] &=\frac{2xy(3y-4)}{(x^2+y^2+1)^2} \\[8px] D_2f(x,y)&=\frac{2(y+2)(x^2+y^2+1)-2y(4x^2+(y+2)^2)}{(x^2+y^2+1)^2}\\[4px] &=\frac{2(-3x^2y+2x^2-2y^2-3y+2)}{(x^2+y^2+1)^2} \end{align} The first derivative vanishes for $x=0$, $y=0$ or $y=4/3$.

We have (the denominator is irrelevant as it doesn't vanish) \begin{align} D_2f(0,y)&=\frac{-2(2y^2+3y-2)}{\dots} \\[4px] D_2f(x,0)&=\frac{4(x^2+1)}{\dots} \\[4px] D_2f(x,4/3)&=\frac{-4(x^2+19/3)}{\dots} \end{align} Thus we only get critical points for $x=0$ and $2y^2+3y-2=0$, that is, $y=-2$ or $y=1/2$.

Since $f(0,-2)=0$ and $f(0,1/2)=5$, the minimum is $0$ and the maximum is $5$.

Well, how do we know that $0$ is the minimum? Because obviously the function only takes on nonnegative values. Why $(0,5)$ is a point of maximum and not a saddle point? Because $f(x,y)$ is bounded: $$ f(x,y)\le\frac{4x^2+4(y+2)^2}{x^2+y^2+1}=4+\frac{16y+12}{x^2+y^2+1}\le16+\frac{16y}{x^2+y^2+1} $$ Now note that $|y|\le x^2+y^2+1$, because $|y|^2-|y|+1\ge0$. Thus $$ |f(x,y)|\le 32 $$ Hence the function must have an absolute maximum which has to be at a critical point. Checking with the Hessian would be indeed a nuisance.