How to find the extreme values of the following function?

Question

Function $$ f(x,y) = 3x^2 + 3xy + y^2 + y^3 $$

My Solution

$$f_x = 6x + 5y + 3y^2 = 0$$

$$f_y = 3x + 2y +3y^2 = 0$$

Solve system of equations to find y=-x, substitute back into $f_x$ and $f_y$ to find (x,y) = (0,0) or/and (-1/3 , 1/3).

$$D = f_{xx} f_{yy} - {f_{xy}}^2$$

This gives that (0,0) is local minimum and (-1/3 , 1/3) is a saddle point. BUT, the actual solution to this question says that (0,0) is the only extreme point, so what happened to (-1/3 , 1/3) ?? Thank you for the help! :)

Answer 1

For this particular function one variable analysis is sufficient. We have $$f(x,y)=3\left (x+{y\over 2}\right )^2++{y^2\over 4}+y^3$$ The function $$g(t,y)=3t^2+{y^2\over 4}+y^3$$ takes the same values as $f(x,y).$ The extreme values of $g(t,y)$ can be determined separately for $t^2$ and for $y^2+4y^3.$ The function $y^2+4y^3$ admits a local minimum at $y=0$ and a local maximum at $y=-{1\over 6}.$ On the other hand the function $t^2$ admits a minimum at $t=0.$ Therefore the function $g(t,y)$ has a local minimum at $(0,0)$ and a saddle point at $(0,-{1\over 6}).$ As $t=x+{y\over 2},$ the function $f$ admits a local minimum at $(0,0)$ and a saddle point at $({1\over 12},-{1\over 6}).$

Answer 2

We have

$$f(x, y) = 3x^2 + 3x y + y^2 + y^3$$

Finding partials

$$\begin{align} f_x &= 6 x + 3 y \\ f_y &= 3 x + 2 y + 3 y^2 \end{align}$$

Setting those to zero and simultaneously solving gives critical points

$$(x, y) = (0, 0), \left(\dfrac{1}{12}, -\dfrac{1}{6} \right)$$

Using the Second Derivative Test, $D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−\big(f_{xy}(x_0,y_0)\big)^2$

• $(0, 0)$ is a local minimum
• $\left(\dfrac{1}{12}, -\dfrac{1}{6} \right)$ is a saddle point