How to optimize $F=(\sqrt{a^2-x^2}-y)^2+(c\sqrt{b^2-y^2}-x)^2, a,b,c>0$

Question

$F=\left(\sqrt{a^2-x^2}-y\right)^2+\left(c\sqrt{b^2-y^2}-x\right)^2,\quad a,b,c>0$

When $c=1$, we can optimize $F$ by using the substitution $x=a \sin(u), y=b \cos(v)$, then $$F=a^2 \cos(u)^2+b^2 \cos(v)^2-2ab \cos(u)\cos(v)+b^2\sin(v)^2 +a^2\sin(y)^2-2ab \sin(v)\sin(u)$$ $$\implies F=a^2+b^2-2ab \cos (u-v).$$ So we get $$F_{min}=(a-b)^2, F_{max}=(a+b)^2. \quad (*)$$

What are other ways of getting (*) and how to do the optimization when $c\ne 1$ ?

Answer 1

I think your way of getting (*) is very nice.

For $c\not=1$, we have $$\begin{align}F_{\text{min}}&=\begin{cases}(a-bc)^2&\text{if $\ \frac{a-bc}{1-c}\leqslant 0$} \\\\0&\text{if $\ 0\lt\frac{a-bc}{1-c}\leqslant b$} \\\\(a-b)^2&\text{if $\ b<\frac{a-bc}{1-c}$}\end{cases} \\ \\\\F_{\text{max}}&=\begin{cases}(a+b)^2&\text{if $\ c<1$} \\\\(a+bc)^2&\text{if $\ 1<c$}\end{cases}\end{align}$$

Proof :

Let $$F(x,y):=\bigg(\sqrt{a^2-x^2}-y\bigg)^2+\bigg(c\sqrt{b^2-y^2}-x\bigg)^2$$ where $-a\leqslant x\leqslant a$ and $-b\leqslant y\leqslant b$.

For every $(X,Y)$ satisfying $X\geqslant 0$ and $Y\geqslant 0$, we have $$F(X,Y)\leqslant F(-X,Y)\leqslant F(-X,-Y)$$ $$F(X,Y)\leqslant F(X,-Y)\leqslant F(-X,-Y)$$

So, from these, we can say the followings :

• To get the minimum, we only need to consider the case where $x\geqslant 0$ and $y\geqslant 0$.

• To get the maximum, we only need to consider the case where $x\leqslant 0$ and $y\leqslant 0$.

We have $$\frac{\partial F}{\partial x}=\frac{2}{\sqrt{a^2-x^2}}\bigg(xy-c\sqrt{b^2-y^2}\sqrt{a^2-x^2}\bigg)$$

So, we get $$\frac{\partial F}{\partial x}=0\iff x=\pm x_0$$ where $$x_0:=\frac{ac\sqrt{b^2-y^2}}{\sqrt{y^2+c^2(b^2-y^2)}}$$

The minimum :

For $0\leqslant x\leqslant a$, we have $F(x,y)\geqslant F(x_0,y)$.

Then, let $$g(y):=F(x_0,y)=\bigg(\sqrt{y^2+c^2(b^2-y^2)}-a\bigg)^2\qquad (0\leqslant y\leqslant b)$$ We get $$g'(y)=\underbrace{\frac{2(1-c^2)^2}{(\sqrt{y^2+c^2(b^2-y^2)}+a)\sqrt{y^2+c^2(b^2-y^2)}}}_{\text{positive}}\bigg(y^2-\frac{a^2-b^2c^2}{1-c^2}\bigg)y$$

Now, we have three cases to consider :

• If $\frac{a^2-b^2c^2}{1-c^2}\leqslant 0$, then since $y^2-\frac{a^2-b^2c^2}{1-c^2}\geqslant 0$, we have $g(y)\geqslant g(0)=(a-bc)^2$.

• If $\frac{a^2-b^2c^2}{1-c^2}\gt b^2$, then since $y^2-\frac{a^2-b^2c^2}{1-c^2}\leqslant 0$, we have $g(y)\geqslant g(b)=(a-b)^2$.

• If $0\lt \frac{a^2-b^2c^2}{1-c^2}\leqslant b^2$, then we have $g(y)\geqslant g\bigg(\sqrt{\frac{a^2-b^2c^2}{1-c^2}}\bigg)=0$.

The maximum :

For $-a\leqslant x\leqslant 0$, we have $F(x,y)\leqslant F(-x_0,y)$.

Then, let $$i(y):=F(-x_0,y)=\bigg(\sqrt{y^2+c^2(b^2-y^2)}+a\bigg)^2\qquad (-b\leqslant y\leqslant 0)$$ We get $$i'(y)=\underbrace{\frac{2(\sqrt{y^2+c^2(b^2-y^2)}+a)}{\sqrt{y^2+c^2(b^2-y^2)}}}_{\text{positive}}(1-c^2)y$$

Now, we have two cases to consider :

• If $1-c^2\gt 0$, then we have $i(y)\leqslant i(-b)=(a+b)^2$.

• If $1-c^2\lt 0$, then we have $i(y)\leqslant i(0)=(a+bc)^2$.

Therefore, we get the conclusion written at the top.$\ \blacksquare$

Answer 2

Hint.

$$ F = \left(a\sqrt{1-\left(\frac xa\right)^2}-b\left(\frac yb\right)\right)^2+\left(c b\sqrt{1-\left(\frac yb\right)^2}-a\left(\frac xa\right)\right)^2 $$

now, as $|\frac xa|\le 1$ and $|\frac yb|\le 1$, calling $\frac xa = \cos u$ and $\frac yb = \cos v$ we have

$$ F = \left(a|\sin u|-b\cos v\right)^2 + \left(c b|\sin v|-a\cos u\right)^2 $$

etc.

Answer 3

Some thoughts: Use Cauchy-Bunyakovsky-Schwarz inequality instead.

We have $$F = a^2 + b^2c^2 + (1 - c^2)y^2 - 2y\cdot \sqrt{a^2 - x^2} - 2c\sqrt{b^2 - y^2}\cdot x.$$

(1) Maximum

By Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} F &\le a^2 + b^2c^2 + (1 - c^2)y^2 + \sqrt{4y^2 + 4c^2(b^2-y^2)}\sqrt{(a^2 - x^2) + x^2}\\ &= a^2 + b^2c^2 + (1 - c^2)y^2 + 2a\sqrt{b^2c^2 + (1 - c^2)y^2}\\ &= \left(a + \sqrt{b^2c^2 + (1 - c^2)y^2}\right)^2. \end{align*}

(2) Minimum

By Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} F &\ge a^2 + b^2c^2 + (1 - c^2)y^2 - \sqrt{4y^2 + 4c^2(b^2-y^2)}\sqrt{(a^2 - x^2) + x^2}\\ &= a^2 + b^2c^2 + (1 - c^2)y^2 - 2a\sqrt{b^2c^2 + (1 - c^2)y^2}\\ &= \left(a - \sqrt{b^2c^2 + (1 - c^2)y^2}\right)^2. \end{align*}

Answer 4

Let $\;u=\sqrt{a^2-x^2},\; v=\sqrt{b^2-y^2},\;$ then the given task about extremes of $$F(x,y)=g(x,y,u,v)=(u-y)^2+(cv-x)^2,$$ gets the conditions $$\begin{cases} u^2+x^2=a^2\\ v^2+y^2=b^2. \end{cases}$$ In accordance with the Lagrange multipliers method, the inner stationary points of the given task can be defined via the unconditional task of six unknowns in the form of $$G(x,y,u,v,\lambda,\mu)=(u-y)^2+(cv-x)^2+\lambda (u^2+x^2-a^2)+\mu(v^2+y^2-b^2),$$ i.e. from the algebraic system $$\begin{cases} (x-cv)+\lambda x=0\\ (y-u)+\mu y=0\\ (u-y)+\lambda u=0\\ c(cv-x)+\mu v=0\\ u^2+x^2=a^2\\ v^2+y^2=b^2.\tag1 \end{cases}$$ $$\begin{cases} u(x-cv)+x(y-u)=0\\ v(y-u)+cy(x-cv)=0, \end{cases}\quad \begin{cases} xy-cuv=0\\ cxy+(1-c^2)yv-uv=0, \end{cases}$$ $$\begin{cases} (1+c)xy+(1-c^2)yv-(1+c)uv=0\\ (1-c)xy-(1-c^2)yv+(1-c)uv=0\\ \end{cases} \begin{cases} xy+(1-c)yv-uv=0\\ xy-(1+c)yv+uv=0\\ \end{cases} \begin{cases} (x-cv)y=0\\ (y-u)v=0\\ \end{cases}$$ $$\left[\begin{align} &cv=x,\qquad u=y,\qquad x^2+y^2=a^2,\quad x^2+c^2y^2=c^2b^2\\ &x=v=0,\quad y^2=b^2\\ &y=u=0,\quad x^2=a^2\\ &y=v=0,\quad 0=b^2,\\ \end{align}\right.$$ $$\left[\begin{align} &x^2=\dfrac{c^2(b^2-a^2)}{c^2-1},\qquad y^2=\dfrac{c^2b^2-a^2}{c^2-1}\\ &x=0,\quad y^2=b^2\\ &y=0,\quad x^2=a^2.\\ \end{align}\right.\tag2$$

System $\,(2)\,$ defines the set of the inner stationary points. Also, the least and the greatest values of $\,F\,$ can be situated at the edges of the area.