$$f(x,y) = x^3 + y^3 - 63(x+y) + 12xy$$ Find maxima and minima of this function.
I tried by taking partial derivative with respect to $x$, which gives $$\frac{∂f}{∂x} = 3(x^2) - 63 + 12y$$ and by taking partial derivative with respect to $y$, $$\frac{∂f}{∂y} = 3(y^2) - 63 + 12x$$
But by equating both equations to $0$, I am unable to get any value for $x$ and $y$ so that I can substitute them in double partial derivative to get maxima and minima. What steps to take ahead?
On
Get rid of the additive constants: $$x^2+4y=y^2+4x\iff x(x-4)=y(y-4).$$ Rule out $x=4$ and $y=0$ as solutions so we can safely divide: $$\frac xy=\frac{y-4}{x-4}.$$ Now use corresponding addition to get $$\frac{x-y}{y}=\frac{y-4-(x-4)}{x-4},$$ that is $$\frac{x-y}{y}=\frac{x-y}{4-x},$$ from which either $x=y$ or $y=4-x$ follows. Now proceed.
On
you have two parabolas, $x^2 + 4 y = 21$ and $y^2 + 4x = 21.$
Add and you have the circle $$ x^2 + y^2 + 4 x + 4 y = 42 $$ or $$ (x+2)^2 + (y+2)^2 = 50 $$
Subtract and you have $x^2 - y^2 - 4 (x-y) = 0,$ or $$ (x-y)(x+y - 4) = 0 $$ which gives two perpendicular lines.
For the line $x=y,$ what do we get for $2(x+2)^2 = 50 \; ? \;$
Finally, you check for maxima with the second partial derivatives. There cannot be any global min or max, and I suspect the local critical points are saddles; I have not yet checked...
Put both derivatives equal to zero: then you have a system $$ 3x^2-63+12y=0 \\ 3y^2-63+12x=0 $$
Isolate $y$ in the first equation and you get $y=-\frac{1}{4}x^2+\frac{21}{4}$. Substitute it in the second equation to get $3(-\frac{1}{4}x^2+\frac{21}{4})^2-63+12x=0$ or equivalently $$\frac{3}{16}x^4-\frac{63}{8}x^2+12x-63+3\cdot \frac{441}{16}=0\\ \frac{3}{16}(x^4-42x^2+48x-336+441)=0\\ \frac{3}{16}(x^4-42x^2+64x+105)=0\\ \frac{3}{16}(x-5)(x-3)(x+1)(x+7)=0$$ Now you can find the corresponding $y$'s and the solutions of the systems are $(-7,-7),(-1,5),(3,3),(5,-1)$ You just need to classify them now