Answer to Question #247892 in Calculus for Susan

Question

Given the function, f(x,y) = x^3 y + 12x^2 - 8y Find all critical points and classify each as a local maximum, local minimum, or neither.

Answer 1

Given function is $f(x,y)=x^3y+12x^2-8y=0$

Differentiating the function w.r.t. $(x),$ we get:

$3x^2y+x^3\frac{dy}{dx}+24x-8\frac{dy}{dx}=0\\ \Rightarrow \frac{dy}{dx}=\frac{24x+3x^2y}{8-x^3}$

For critical points, put $\frac{dy}{dx}=0,$ we get:

$\frac{24x+3x^2y}{8-x^3}=0\\ \Rightarrow 24x+3x^2y=0\\ \Rightarrow 3x(8+xy)=0$

$\therefore x=0 \ or \ xy=-8$

Case 1: If $x=0$ , then:

$f(x,y)=-8y=0\\ \Rightarrow y=0$

Case 2: If $xy=-8$ , then:

$f(x,y)=-8x^2+12x^2-8y=0\\ \Rightarrow 4x^2-8y=0\\ \Rightarrow \frac{x^2}{2}=y$

So, the given function can be written as:

$x^3(\frac{x^2}{2})+12x^2-4x^2=0\\ \Rightarrow \frac{x^5}{2}+8x^2=0\\ \Rightarrow x^2(\frac{x^3}{2}+8)=0\\ \Rightarrow x=0 \ or\ x^3=-16\\ \Rightarrow x=0 \ or\ x=-(16)^{\frac{1}{3}}$

$\therefore y=0 \ or\ y=\frac{(16)^{\frac{2}{3}}}{2}$

So, from both cases, we obtain the same point: $(0,0)$ and $(-(16)^{\frac{1}{3}}, \frac{(16)^{\frac{2}{3}}}{2})$

Now, calculating double derivative, we get:

$6xy+3x^2\frac{dy}{dx}+3x^2\frac{dy}{dx}+x^3\frac{d^2y}{dx^2}+24-8\frac{d^2y}{dx^2}=0\\ \Rightarrow \frac{d^2y}{dx^2}=\frac{6xy+6x^2\frac{dy}{dx}+24}{8-x^3}$

Now, calculating value of double derivative at $(0,0)$ , we get:

$(\frac{d^2y}{dx^2})_{(0,0)}=\frac{24}{8}=3$ which is positive. Hence, the point is local minimum.

Now, calculating value of double derivative at $(-(16)^{\frac{1}{3}}, \frac{(16)^{\frac{2}{3}}}{2})$, we get:

$(\frac{d^2y}{dx^2})_{(-(16)^{\frac{1}{3}}, \frac{(16)^{\frac{2}{3}}}{2})}=\frac{-48+0+24}{8+16}=-1$ which is negative. Hence, the point is local maximum.