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Consider the function f(x,y) = y\sqrt{x} - y^2 - 1x + 3y. Find and classify all critical points...

Question:

Consider the function {eq}\, f(x,y) = y\sqrt{x} - y^2 - 1x + 3y {/eq}.

Find and classify all critical points of the function.

Critical point on a two variable function

This example illustrates how to find the coordinates of the critical points on a two variable function using partial differentiation.

The example goes on to show how to classify the critical points as maxima, minima, or saddle points.

Answer and Explanation:

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

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

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

To classify the critical points, we use {eq}\displaystyle D = \frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2} - \left (\frac{\partial^2 f}{\partial x\partial y} \right )^2 {/eq}.

We evaluate D at the critical point.

The type of critical point is classified as follows:

(1) {eq}\displaystyle D > 0, \: \frac{\partial^2 f}{\partial x^2} > 0 {/eq} - the critical point is a local (or relative) minimum.

(2) {eq}\displaystyle D > 0, \: \frac{\partial^2 f}{\partial x^2} < 0 {/eq} - the critical point is a local (or relative) maximum.

(3) {eq}\displaystyle D < 0 {/eq} - the critical point is a saddle point.

(4) {eq}\displaystyle D = 0 {/eq} - the critical point could be a local maximum, or a local.minimum, or a saddle point. Other more complex techniques will need to be used to determine which it is.

In this example we have {eq}f(x,y) = y\sqrt{x} - y^2 - 1x + 3y = yx^{\frac{1}{2}} - y^2 - x + 3y {/eq}.

First of all, find the partial derivatives that we're going to need:

{eq}\displaystyle \frac{\partial f}{\partial x} = \frac{1}{2}yx^{-\frac{1}{2}} - 1 \\ \displaystyle \frac{\partial f}{\partial y} = x^{\frac{1}{2}} - 2y + 3 \\ \displaystyle \frac{\partial^2 f}{\partial x^2} = -\frac{1}{4}yx^{-\frac{3}{2}} \\ \displaystyle \frac{\partial^2 f}{\partial y^2} = -2 \\ \displaystyle \frac{\partial^2 f}{\partial x\partial y} = \frac{1}{2}x^{-\frac{1}{2}} {/eq}

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

So

{eq}\displaystyle (1)\: \frac{1}{2}yx^{-\frac{1}{2}} - 1 = 0 \\ (2)\: x^{\frac{1}{2}} - 2y + 3 = 0 {/eq}.

{eq}\displaystyle (1) \Rightarrow yx^{-\frac{1}{2}} = 2 \Rightarrow y = 2x^{\frac{1}{2}}. {/eq}

Substitute for y in (2):

{eq}x^{\frac{1}{2}} - 2(2x^{\frac{1}{2}}) + 3 = 0 \\ \Rightarrow 3x^{\frac{1}{2}} = 3 \\ \Rightarrow x^{\frac{1}{2}} = 1 \\ \Rightarrow x = 1. {/eq}

Substitute {eq}x = 1 {/eq} in (2):

{eq}1^{\frac{1}{2}} - 2y + 3 = 0 \\ \Rightarrow y = 2. {/eq}

So there is only one critical point, at {eq}(1, 2). {/eq}

To classify the critical point, we need

{eq}\displaystyle D = \frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2} - \left (\frac{\partial^2 f}{\partial x\partial y} \right )^2 \\ \displaystyle = (-\frac{1}{4}yx^{-\frac{3}{2}})(-2) - (\frac{1}{2}x^{-\frac{1}{2}})^2 {/eq}

At the point (1, 2), we have

{eq}\displaystyle D = (-\frac{1}{4}(2)(1)^{-\frac{3}{2}})(-2) - (\frac{1}{2}(1)^{-\frac{1}{2}})^2 = \frac{3}{4} {/eq}

Since {eq}D > 0 {/eq}, this is a minimum point.


Learn more about this topic:

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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
195K

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