# Consider the following function. f(x,y)=(x-3) \ln(x^5y) A) Find the critical points of f. B)...

## Question:

Consider the following function.

{eq}f(x,y)=(x-3) \ln(x^5y) {/eq}

A) Find the critical points of {eq}f {/eq}.

B) Using your critical points in A), find the value of {eq}D(a,b) {/eq} from the second partials test that is used to classify the critical points.

C) Use the second partials test to classify the critical points from A).

## Critical points of a two variable function

This example illustrates the process involved in finding the critical points of a function of two variables.

It also examines how to classify the critical points by using the second partial differentials.

Part (A)

Suppose we have a function {eq}f(x,y) {/eq}.

Then, the critical points (which may be local (or relative) maxima and minima) of the function will occur when

both {eq}\dfrac{\partial f}{\partial x} = 0, \: \dfrac{\partial f}{\partial y} = 0 {/eq} simultaneously.

In the current example: {eq}f(x, y) = (x - 3) \ln(x^5y) = (x - 3) (5\ln x + \ln y). {/eq}

Finding the first order partial derivatives:

{eq}\dfrac{\partial f}{\partial x} = 5\ln x + \ln y + \dfrac{5(x - 3)}{x} \\ \dfrac{\partial f}{\partial y} = \dfrac{x - 3}{y} {/eq}

At a critical point, {eq}\dfrac{\partial f}{\partial x} = 0, \: \dfrac{\partial f}{\partial y} = 0 {/eq} simultaneously.

So

{eq}(1)\: 5\ln x + \ln y + \dfrac{5(x - 3)}{x} = 0 \\ (2)\: \dfrac{x - 3}{y} = 0 {/eq}.

Now

{eq}(2) \Rightarrow x = 3. {/eq}

Then put this x value in (1):

{eq}5\ln 3 + \ln y + \dfrac{5(3 - 3)}{x} = 0 \\ \Rightarrow \ln y = -5\ln 3 = \ln(3^{-5}) \\ \Rightarrow y = 3^{-5} = \dfrac{1}{243} {/eq}

So there is only one critical point, at {eq}(3, \dfrac{1}{243}). {/eq}

Part B

The second partials test for the type of critical point uses

{eq}D = \dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial^2 f}{\partial y^2} - \left (\dfrac{\partial^2 f}{\partial x\partial y} \right )^2 {/eq}.

The second partial derivatives are as follows:

{eq}\dfrac{\partial^2 f}{\partial x^2} = \dfrac{5}{x} + \dfrac{15}{x^2} \\ \dfrac{\partial^2 f}{\partial y^2} = (x - 3)(-y^{-2}) \\ \dfrac{\partial^2 f}{\partial x\partial y} = \dfrac{1}{y} {/eq}

Hence

{eq}D = \dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial^2 f}{\partial y^2} - \left (\dfrac{\partial^2 f}{\partial x\partial y} \right )^2 \\= (\dfrac{5}{x} + \dfrac{15}{x^2})(x - 3)(-y^{-2}) - (\dfrac{1}{y})^2 {/eq}

Using the notation given in the question:

{eq}D(a, b) = (\dfrac{5}{a} + \dfrac{15}{a^2})(a - 3)(-b^{-2}) - (\dfrac{1}{b})^2 {/eq}

Part C

At the critical point {eq}(3, \dfrac{1}{243}), {/eq} we have

{eq}D = (\dfrac{5}{3} + \dfrac{15}{3^2})(3 - 3)(-(\dfrac{1}{243})^{-2}) - \left(\dfrac{1}{\dfrac{1}{243}}\right)^2 = 59049. {/eq}

Also at the critical point:

{eq}\dfrac{\partial^2 f}{\partial x^2} = \dfrac{5}{x} + \dfrac{15}{x^2} = \dfrac{5}{3} + \dfrac{15}{3^2} = \dfrac{10}{3} {/eq}

Since {eq}D > 0, \: \dfrac{\partial^2 f}{\partial x^2} > 0, {/eq} the critical point is a local minimum. 