## Problem on the second derivative test

Let $A(x,y) = 2 x y + \frac{64}{y} + \frac{32}{x}$. Show that $x=2, y=4$ is a critical point. Verify that the point is a local minimum of $A$.
• ## Solution

#### Showing the critical point

Recall that
Hence, we compute both partial derivatives of $A(x,y)$:
Plugging in $x=2, y=4$, we verify both partial derivatives are 0.

#### Verifying the critical point is a local minimum

To verify $x=2, y=4$ is a minimizer, we recall the second derivative test:
The second derivatives of $A$ are \begin{align} A_{x x} &= \frac{64}{x^3} \\ A_{x y} = A_{y x} &= 2 \\ A_{y y} &= \frac{128}{y^3} \end{align}
Evaluating at $x=2, y=4$ gives $A_{x x} = 8, A_{x y} = 2, A_{y y} = 2$
Because this determinant is positive and $A_{xx} >0$, the point $x=2, y=4$ is a local minimizer of $A$.