$$f(x,y) = x^3 + y^3 – 63(x+y) + 12xy$$
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?
Best Answer
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