Find the absolute extrema of the function $f(x,y)=x^2 + xy + y^2$ on the closed disk $x^2 + y^2 \leq 1$
I found $f_x = 2x + y $
$f_y= 2y+x$
Equated them to zero and Solving them gives $x=0 \; \; y=0$
But I am having difficulty finding critical points on the boundary
If I solve for $x \; or \; y$ in the disk equation, I will get $\pm$ values
How can I get the critical points on the boundary ?
On
$$f(x,y)=x^2+xy+y^2 = \dfrac{1}{2}((x+y)^2+x^2+y^2)\le \dfrac{1}{2}(x^2+y^2+2(x^2+y^2))\le \dfrac{3}{2}$$
The extrema occurs at $(x,y)=(\pm \frac{1}{\sqrt{2}},\pm \frac{1}{\sqrt{2}})$
On
Have a look at my answer in this post : $\sin\alpha+\cos\alpha<2$
On the boundary, let's consider $(x,y)$ in the first quadrant: $$f(x,y)=x^2+y^2+xy=1+xy=1+\mathcal A$$
Where $\mathcal A$ is the area of the rectangle determined by $x=\cos(\theta)$ and $y=\sin(\theta)$.
The area of a rectangle with fixed diagonal (here $1$ the radius of the circle) is maximum for a square.
So $f(x,y)=1+\mathcal A$ is maximum when $x=y=\frac1{\sqrt{2}}$ or $\theta=\frac{\pi}4$.
The other solution is $(-x,-y)$ by symmetry.
On
Use symmetry of the objective function and the form of the constraint to notice that the maximum occurs where the line $x=y$ intersects the unit circle. Then the maximized value is $3/2$.
On
Use polar coordinates. $$x^2+xy+y^2=r^2(\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta)=r^2\left(1+\frac{1}{2}\sin(2\theta)\right)$$ With $\theta$ varying from $0$ to $2\pi$, the maximum of the expression in parentheses is $3/2$ at $2\theta=\pi/2$, and the minimum is $1/2$. Since $r^2$ varies between $0$ and $1$, your function has a minimum of $0$ at $(0,0)$, and a maximum of $3/2$ at $x=y=\frac{1}{\sqrt 2}$
By Lagrange's multipliers,
$$h(x,y)=x^2+xy+y^2+\lambda (x^2+y^2)$$
$$h_x=0\implies2x+y+\lambda(2x)=0$$
$$h_y=0\implies2y+x+\lambda(2y)=0$$
Now, write $x$ and $y$ in terms of $\lambda$ and substitute in $x^2+y^2=1$ to get $\lambda$ and then the values of $(x,y)$