Help with a partial derivative and stationary point question (almost complete)?

Question

Hi I'm stuck on this question which I have attached a picture of below along with my working. I have the answers and so I know I have calculated the correct x and y values but when I sub this into the formula for D (second derivative test formula) I get -8 and 8 which doesn't match the answer given to me which is -7/12 saddle point and -45/4 local minimum. I'm sure I have done the partial derivatives correct also so I'm very confused and hoping someone could help me. Thank you so much

Answer 1

You are given that $$g(x,y)=\frac{x^3}{3}+y^2-2xy-3y$$ So, the partial derivatives are

$$g_x=x^2-2y$$ $$g_{xx}=2x$$ $$g_y=2y-2x-3$$ $$g_{yy}=2$$ $$g_{xy}=-2$$

If we set $g_x=0$ then

$$g_x=x^2-2y=0 \implies x^2=2y$$

where

$$x=\pm \sqrt{2y}\tag{1}$$ $$y = \frac{x^2}{2}\tag{2}$$

and if we set $g_y=0$ then

$$g_y=2y-2x-3=0 \implies 2(y-x)=3 \implies y-x = \frac{3}{2}$$

where

$$y=x+\frac{3}{2}\tag{3}$$ $$x=y-\frac{3}{2}\tag{4}$$

So, equating (1) and (4) forms

$$y-\frac{3}{2}=\pm\sqrt{2y}$$

which can be reduced to

$$y=\frac{9}{2},~~~y=\frac{1}{2}$$

which substituting back into $(2)$ forms

$$\frac{9}{2}=\frac{x^2}{2} \implies x=3,-3$$ $$\frac{1}{2}=\frac{x^2}{2} \implies x=1,-1$$

which gives us the critical points $(-1,\frac{1}{2}), (1,\frac{1}{2}),(3,\frac{9}{2})$ and $(-3,\frac{9}{2})$.

We now need to apply the second partial derivative test. The test tells us that provided $g(x, y)$ is a differentiable real function of two variables whose second partial derivatives exist we can compute

$$D(x,y)=g_{xx}(x,y)g_{yy}(x,y)-\Big(g_{xy}(x,y)\Big)^2$$

and that

1. If $D(a,b)>0$ and $g_{xx}(a,b)>0$ then $(a,b)$ is a local minimum of $g$.
2. If $D(a,b)>0$ and $g_{xx}(a,b)<0$ then $(a,b)$ is a local maximum of $g$.
3. If $D(a,b)<0$ then $(a,b)$ is a saddle point of $g$.
4. If $D(a,b)=0$ then the second derivative test is inconclusive, and the point $(a, b)$ could be any of a minimum, maximum or saddle point.

We found before that

$$g_{xx}=2x$$ $$g_{yy}=2$$ $$g_{xy}=-2$$

Thus, if we compute $D(x,y)$ for our four critical points we have

$$D\big(-1,\frac{1}{2}\big)=(2\times -1)(2)-(-2)^2=-8 <0$$ $$D\big(1,\frac{1}{2}\big)=(2\times 1)(2)-(-2)^2=0$$ $$D\big(3,\frac{9}{2}\big)=(2\times 3)(2)-(-2)^2=8>0$$ $$D\big(-3,\frac{9}{2}\big)=(2\times -3)(2)-(-2)^2=-16<0$$

Therefore, by the second partial derivative test,

1. $(-1,\frac{1}{2})$ is a saddle point by definition.
2. $(1,\frac{1}{2})$ is inconclusive.
3. $(3,\frac{9}{2})$ is a local minimum since $D\big(3,\frac{9}{2}\big) > 0$ and $g_{xx}(3,\frac{9}{2})=6>0$.
4. $(-3,\frac{9}{2})$ is a saddle point by definition.

To get the value of $-\frac{45}{4}$, evaluate $g(x,y)$ at $(x,y)=(3,\frac{9}{2})$ to form

$$g\big(3,\frac{9}{2}\big)=\frac{3^3}{3}+\big(\frac{9}{2}\big)^2 - 2(3)\big(\frac{9}{2}\big)-3\big(\frac{9}{2}\big)=-\frac{45}{4}$$

which is the local minimum that you found through Wolfram Alpha. Similarly, to get $\frac{-7}{12}$ evaluate $g(x,y)$ at $(x,y)=(-1,\frac{1}{2})$ to form

$$g\big(-1,\frac{1}{2}\big)=\frac{(-1)^3}{3}+\big(\frac{1}{2}\big)^2 - 2(-1)\big(\frac{1}{2}\big)-3\big(\frac{1}{2}\big)=-\frac{7}{12}$$

which is a saddle point.