Problem on the second derivative test - Leading Lesson

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
Critical points of functions of several variables

The critical points of a function are points where all of the partial derivatives are zero.
Hence, we compute both partial derivatives of $A(x,y)$:
Partial derivatives

$$\partial_x A = 2 y - \frac{32}{x^2}, \qquad \partial_y A = 2x - \frac{64}{y^2}.$$
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:
Second derivative test for functions of several variables

Let $D = \text{det } \begin{pmatrix} A_{x x} & A_{x y}\\ A_{y x} & A_{y y} \end{pmatrix} = A_{x x} A_{y y} - A_{x y}^2 $.
If $D > 0$ and $A_{x x} > 0$ at a critical point, then it is a minimum.
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 $A_{x x} = 8, A_{x y} = 2, A_{y y} = 2$
Determinant of a 2x2 matrix

The second derivative matrix has determinant $$\text{det } \begin{pmatrix}8 & 2\\ 2 & 2 \end{pmatrix} = 12$$
Because this determinant is positive and $A_{xx} >0$, the point $x=2, y=4$ is a local minimizer of $A$.

Related topics

Multivariable calculus
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