Question

Find the absolute maximum and minimum values of the function f on the set

Answer 1

$f_x=4y^2-2xy^2-y^3=y^2(4-2x-y)=0\implies y=0\text{ or }2x+y=4$

$f_y=8xy-2x^2y-3xy^2=xy(8-2x-3y)=0\implies x=0\text{ or }y=0\text{ or }2x+3y=8$

For now, ignore any critical points that occur on the boundary of $\mathcal D$.

Case 1: If $y=0$, then either $x=0$ or $2x=8\implies x=4$, so we have two critical points (0, 0) and (4, 0), both on the boundary.

Case 2: Suppose $2x+y=4$. Then ...

Case 2a: ... if $x=0$, then $y=4$ - critical point at (0, 4) on the boundary.

Case 2b: ... if $y=0$, then $2x=4\implies x=2$ - critical point at (2, 0) on the boundary.

Case 2c: ... if $2x+3y=8$, then

$\begin{cases}2x+y=4\\2x+3y=8\end{cases}\implies x=1\text{ and }y=2$

- critical point at (1, 2) not on the boundary.

We compute the Hessian matrix of $f(x,y)$:

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}-2y^2&4(2-x)y-3y^2\\4(2-x)y-3y^2&8x-2x^2-6xy\end{bmatrix}$

$\det\mathbf H(1,2)=32>0\text{ and }f_{xx}(1,2)=-8$

which means (1, 2) is a local maximum, where we get $f(1,2)=4$.

Now we consider the boundary in three parts:

(1) Fix $x=0$. Then $f(0,y)=0$. It's possible that 0 is the smallest value of $f(x,y)$ whereever you go.

(2) Fix $y=0$. Then $f(x,0)=0$, and we get the same result as in the previous case.

(3) Fix $x+y=6$. Then $f(x,6-x)=-2x(x-6)^2$. Denote this function by $F(x)$. We compute the first derivative and find the critical points:

$F'(x)=-2(x-6)^2-4x(x-6)=-2(x-6)(3x-6)=0\implies x=6\text{ or }x=2$

then compute the second derivative and determine its sign at these points:

$F''(x)=-2(3x-6)-2(x-6)(3)=-12(x-4)\implies\begin{cases}F''(6)=-240\end{cases}$

which indicates that a maximum occurs as $x=6$ and a minimum at $x=2$. Respectively, we have $6+y=6\implies y=0$ and $2+y=6\implies y=4$, and $f(6,0)=0$ and $f(2,4)=-64$.

So the absolute maximum of $f(x,y)$ over $\mathcal D$ is 4 at (1, 2), and the absolute minimum is -64 at (2, 4).