$x^2 + y^2+xy = 1$ , then find the minimum of $x^3 y + xy^3 +4$

Question

x and y belongs to real numbers. $ x^2 + y^2+xy = 1 $. then find the minimum value of $x^3 y + xy^3 +4$.
I assume $ x = r \sin (w)$ and $ y = r\cos(w) $. $ x^3 y + xy^3 +4 = L $ which give me $ \frac{2}{3} \le r^2 \le 2 $ I am stuck after that.Its my Humble request to help me after that.

Answer 1

Hint: $x^2+y^2=1-xy$, so that $x^3y+xy^3+4=xy(x^2+y^2)+4=xy(1-xy)+4$. Substitute $z=xy$. What do you end up with?

Answer 2

For $x=1$ and $y=-1$ we'll get a value $2$.

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

Indeed, we need to prove that $$x^3y+y^3x+4\geq2$$ or $$x^3y+y^3x+2(x^2+xy+y^2)^2\geq0$$ or $$2x^4+5x^3y+6x^2y^2+5xy^3+2y^4\geq0$$ or $$2x^4+4x^3y+2x^2y^2+x^3y+2x^2y^2+xy^3+2x^2y^2+4xy^3+2y^4\geq0$$ or $$(x+y)^2(2x^2+xy+2y^2)\geq0.$$ Done!

Answer 3

Lagrangian multiplier for constrained function

$$L(x,y)= F(x,y)-\lambda G(x,y) $$

set in the order given, partially differentiating with respect to both variables

$$ \dfrac{F_x}{F_y}=\dfrac{G_x}{G_y}= \lambda $$

$$\dfrac{2x+y}{2y+x}=\dfrac{3x^2y+x^3}{x^3+3y^2} $$

Cross multiplying to simplify we get two conditions

$$ (x^2-y^2)(x^2+y^2-xy)=0 $$

Minimal value verifiable by sign of second partial derivatives of $L(x,y)$..

Answer 4

Your method of solution is good. You just need to continue. By setting $x=r\cos\theta$ and $y=r\sin\theta$, we get$$\cos\theta\sin\theta=\frac1{r^2}$$ and so $\tfrac23\leqslant r^2\leqslant2$, as you found. We have $$x^3y+xy^3+4=r^2-r^4+4=\tfrac{17}4-(r^2-\tfrac12)^2,$$which is minimum when $|r^2-\frac12|$ is maximum, namely when $r^2=2$.