Exercise 9.21

(a) We have $D_{1}f(x,y)=6x(x-1)$ so $D_{1}f(x,y)$ if $x=0$ or $x=1$. Also, $D_{2}f(x,y)=6y(6+1)$ so $D_{2}f(x,y)=0$ if $y=0$ or $y=-1$. Hence the gradient of $f$ equals 0 at the four points $(0,0)$, $(1,0)$, $(0,-1)$, and $(1,-1)$. To tell whether these are local maxima or minima, or saddle points, the easiest way is to apply the second derivative test for multivariable functions, which involves finding the eigenvalues of the matrix of second derivatives (the “Hessian”) at these points, but that wasn’t demonstrated in the text, so we need to show this more directly.

At $(1,0)$, for small values $d_1$ and $d_2$, we have

which is positive for small values of $d_1$ and $d_2$. Hence $f$ has a local minimum at $(1,0)$.

Similarly, at $(0,-1)$ we have

which is negative for small values of $d_1$ and $d_2$. Hence $f$ has a local maximum at $(1,0)$.

At $(0,0)$ $f(x,0)=2x^3-3x^2$ can be shown to have a local maximum at $x=0$, using the usual calculus techniques. However, $f(0,y)=2y^3+3y^2$ has a local minimum at $y=0$. Hence $(0,0)$ is a saddle point for $f$.

Similarly, at $(1,-1)$ $f(x,-1)=2x^3-3x^2+1$ has a local minimum at $x=1$, but $f(1,y)=2y^3+3y^2-1$ has a local maximum at $y=-1$. Hence $(1,-1)$ is also a saddle point for $f$.

(b) Note that

The first factor shows that $f$ is equal to zero along the diagonal $y=-x$. The zero set of the second factor of degree 2 must be some sort of conic section. Converting to polar coordinates we get

which can be seen to describe a ellipse symmetric with the diagonal line $y=-x$ and intersecting it at the points $(0,0)$ and $(1,-1)$.

The points of the zero set along the diagonal satisfy the relation $y=-x$ except at the two intersection points, where we cannot describe $x$ or $y$ as single-valued functions of each other.

Let $g(x,y)=2(x^2-\mathit{xy}+y^2)-3(x-y)$. Along the zero set of $g(x,y)$, $x$ cannot be expressed as a function of $y$ where $D_{1}g(x,y)=4x-2x-3=0$. (I am using the elliptical zero set as described above; the general case is more complicated.) That is, we are looking for the intersection of the zero set of $g$ with the line $y=2x-(3∕2)$, which occurs at the points $(0,-3∕2)$ and $(1,1∕2)$.

Similarly, $y$ cannot be expressed as a function of $x$ where $D_{2}g(x,y)=-2x+4y+3=0$, so we are looking for the intersection of the zero set of $g$ with the line $x=2y+(3∕2)$, which occurs at the points $(3∕2,0)$ and $(-1∕2,-1)$.