Finding minimum and maximum of 2-variable function in a closed area

Question

I have this function: $f(x,y)=2x^3+3x^2y-3y$

I need to find the absolute minimum and maximum points that are within the triangle that is set by: $y=\frac{1}{3}x, x=1, y=0$

I started like this:

$$f_x(x,y) = 6x^2+6xy = 0$$ $$f_y(x,y) = 3x^2-3 = 0$$

Got the points: $(-1,1),(1,-1)$

But if we look at the intersection of all the lines that create the triangle, the points must be in $0\le x\le1$ and $0\le y\le \frac{1}{3}$

So how do I continue from here?

Thanks.

Answer 1

EDIT: This is the solution for the originally posted function $f(x,y)=2x^3+3x^2y-3y$. But the principles are the same.

Looking at the partial derivatives shows that there is no local mininum or maximum inside the triangle.

Now let us look at the border, i.e. at the triangle sides and corners:

1. side $y=\frac{1}{3}x$: $f_1(x) = 2x^3+2x^2-x$, $f_1'(x) = 6x^2+4x-1 = 0$ has a root in $[0,1]$: $x_0 = \frac{\sqrt{\frac{5}{2}}-1}{3} = 0.19\ldots$, $f_1(x_0) = \frac{13-5 \sqrt{10}}{27} = -0.10\ldots$
2. side $x=1$: $f_2(y) = 4-3y$ has no local extremum
3. side $y=0$: $f_3(x) = 2x^3+2x^2$, $f_1'(x) = 6x^2+4x$ has roots at $x_1=0, x_2=\frac{2}{3}$, $f_3(x_1)=0, f_3(x_2)=\frac{40}{27}=1.48\ldots$
4. corner $(0,0)$: $f(0,0)=0$
5. corner $(1,0)$: $f(1,0)=4$
6. corner $(1,\frac{1}{3})$: $f(1,\frac{1}{3}) = 1$

So all together we have

• the minimum $f(x_0,\frac{1}{3}x_0) = f(0.19\ldots,0.06\ldots) = \frac{13-5\sqrt{10}}{27} = -0.10\ldots$ and
• the maximum $f(1,0)=4$.