Partial Derivatives : Given $f(x) = Ax^3 + By^3 - Cx - Dy + E$

Question

Given $f(x) = Ax^3 + By^3 - Cx - Dy + E$
Propose any value for $A, B, C, D$ and $E$ so that these will give three(3) critical points.
Please help. My choice of value only gives me two(2) critical point therefore it is not accurate.
I randomly pick value for $A, B, C, D$ and $E$, then I differentiate the equation.
I find $frac{partial}{partial x} (Ax^3 + By^3 - Cx - Dy + E )$ and
$frac{partial}{partial y}(Ax^3 + By^3 - Cx - Dy + E )$.
Both equation that I differentiated is then equal = $0$
Then I determine the value of $x$ and $y$ which will give me the critical points
p/s: Very sorry I did not know how to use the math format

miracle173 · Answer

To calculate the critical points of the function
$$f(x,y) = Ax^3 + By^3 - Cx - Dy + E$$
you have to find appropriate values $A,ldots,F$ where
$$frac partial {partial x}f=0$$
$$frac partial {partial y}f=0$$
, so
$$3Ax^2-C=0$$
$$3By^2-D=0$$
The solutions of these quadradic equations are independent of each other. The first quadratic equation has two real solutions if the C is not $0$ and $A$  has the same sign as $C$. If $A=C=0$ we have infinitely many solutions.
If  $x_1$ is a solution of the first euation and $y_1$ is a solution of the second one, then $(x_1,y_1)$ is a solution of both equations.
So if we choose $A=B=1$, $C=D=3$, $E=1$ then the 4 critical points are $(1,1),(1,-1),(-1,1),(-1,-1)$

Partial Derivatives : Given $f(x) = Ax^3 + By^3 - Cx - Dy + E$

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