minimum value of of $(5+x)(5+y)$

Question

If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$

Then find the minimum value of $(5+x)(5+y)$

What I try

$$(5+x)(5+y)=25+5(x+y)+xy$$

$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$

I am finding $f(x,y)=22+5(x+y)+(x+y)^2$

How do I solve it? Help me please

Answer 1

The feasible set is a (compact) ellipse $E$ in the $(x,y)$-plane. We have to consider the Lagrangian $$\Phi:=(x+5)(y+5)-\lambda(x^2+xy+y^2-3)\ .$$ The equations $$\Phi_x=y+5-\lambda(2x+y)=0,\qquad\Phi_y=x+5-\lambda(x+2y)=0\ ,$$ or $$\left\{\eqalign{2\lambda x+(\lambda-1)y&=5 \cr (\lambda-1)x+2\lambda y&=5\cr}\right.$$ are symmetric in $x$ and $y$, hence the solution is given by $$x=y={5\over3\lambda-1}\ .\tag{1}$$ Plugging this into the equation $x^2+xy+y^2=3$ of $E$ gives $(3\lambda-1)^2=25$, or $3\lambda-1=\pm5$. From $(1)$ we then get two conditionally stationary points of $f$ on $E$, namely $(1,1)$ and $(-1,-1)$. From $f(1,1)=36$ and $f(-1,-1)=16$ it then follows that the minimum of $f$ on $E$ is $16$.

Answer 2

Hint: From the equation $$y^2+xy+x^2-3=0$$ we get $$y_{1,2}=-\frac{x}{2}\pm\sqrt{3-\frac{3}{4}x^2}$$ so you will obtain the two functions $$f_1(x)=(5+x)\left(5-\frac{x}{2}+\sqrt{3-\frac{3}{4}x^2}\right)$$ or $$f_2(x)=(5+x)\left(5-\frac{x}{2}-\sqrt{3-\frac{3}{4}x^2}\right)$$ to consider. Doing this we get $$(5+x)(5+y)\geq 16$$, the equal sign holds if $$x=y=-1$$ as Mr. Rozenberg stated.

Answer 3

Let $x=y=-1$.

Thus, we obtain the value $16$.

We'll prove that it's a minimal value.

Indeed, $$(5+x)(5+y)-16=xy+5(x+y)+9=3x^2+4xy+3y^2+5(x+y)\sqrt{\frac{x^2+y^2+xy}{3}}\geq$$ $$\geq3x^2+4xy+3y^2-5\sqrt{\frac{(x+y)^2(x^2+y^2+xy)}{3}}=$$ $$=\frac{3(3x^2+4xy+3y^2)^2-25(x+y)^2(x^2+y^2+xy)}{\sqrt3(\sqrt{3}(x^2+y^2+xy)+5\sqrt{(x+y)^2(x^2+y^2+xy)})}=$$

$$=\frac{(x-y)^2(2x^2+xy+2y^2)}{\sqrt3(\sqrt{3}(x^2+y^2+xy)+5\sqrt{(x+y)^2(x^2+y^2+xy)})}\geq0$$ and we are done!

Answer 4

The symmetry of the problem suggests to substitute $x = u+v$, $y = u-v$. Then the given constraint is $$ 3 = x^2 + xy + y^2 = 3u^2 + v^2 $$ and in particular $-1 \le u \le 1$.

We look for the minimal value of $$ (5+x)(5+y) = (5+u)^2 - v^2 = (5+u)^2 - (3-3u^2) \\ = 4u^2 + 10 u + 22 \, . $$ This expression is increasing for $u \in [-1, 1]$, and therefore $$ \ge 4 (-1)^2 + 10(-1) + 22 = 16 \, . $$ Equality holds for $u=-1, v=0$, corresponding to $x=-1, y=-1$.

Answer 5

Using Rotation of axes

if $t$ is the rotation of axes, $\cot2t=0\iff\cos2t=0$

Set $2t=\dfrac\pi2$

$\sqrt2u=x+y,\sqrt2v=y-x$

$$3=\dfrac{(u-v)^2+(u+v)^2}2=u^2+v^2$$

WLOG $u=\sqrt3\cos t, v=\sqrt3\sin t$

$$(5+x)(5+y)=25+\sqrt2u+\dfrac{2(u^2-v^2)}4=\dfrac{50+2\sqrt2u+u^2-v^2}2$$

$2\sqrt2u+u^2-v^2=2\sqrt2(\sqrt3\cos t)+3(\cos^2t-\sin^2t)$

$=6\cos^2t+2\sqrt6\cos t-3=6\left(\cos t+\dfrac1{\sqrt6}\right)^2-3-6\left(\dfrac1{\sqrt6}\right)^2$

Now $-1\le\cos t\le1\iff -1+\dfrac1{\sqrt6}\le\cos t+\dfrac1{\sqrt6}\le1+\dfrac1{\sqrt6}$

$\implies\left(\cos t+\dfrac1{\sqrt6}\right)^2\le\left(1+\dfrac1{\sqrt6}\right)^2$