How To Find Relative Extrema

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Department 3-three : Relative Minimums and Maximums

In this department nosotros are going to extend ane of the more of import ideas from Calculus I into functions of two variables. We are going to start looking at trying to find minimums and maximums of functions. This in fact will exist the topic of the following ii sections every bit well.

In this section we are going to exist looking at identifying relative minimums and relative maximums. Call up as well that we will often use the word extrema to refer to both minimums and maximums.

The definition of relative extrema for functions of two variables is identical to that for functions of one variable we simply demand to remember now that nosotros are working with functions of two variables. So, for the sake of completeness here is the definition of relative minimums and relative maximums for functions of two variables.

Definition

A function $f\left( {x,y} \right)$ has a relative minimum at the point $\left( {a,b} \right)$ if $f\left( {x,y} \correct) \ge f\left( {a,b} \right)$ for all points$\left( {10,y} \right)$ in some region effectually $\left( {a,b} \right)$.
A function $f\left( {x,y} \right)$ has a relative maximum at the betoken $\left( {a,b} \right)$ if $f\left( {x,y} \right) \le f\left( {a,b} \right)$ for all points$\left( {10,y} \right)$ in some region effectually $\left( {a,b} \correct)$.

Notation that this definition does not say that a relative minimum is the smallest value that the function volition ever take. It only says that in some region around the point $\left( {a,b} \right)$ the function will always be larger than $f\left( {a,b} \correct)$. Outside of that region information technology is completely possible for the function to be smaller. Likewise, a relative maximum just says that around $\left( {a,b} \right)$ the office will ever be smaller than $f\left( {a,b} \right)$. Over again, exterior of the region it is completely possible that the function will exist larger.

Side by side, we need to extend the thought of critical points upward to functions of two variables. Remember that a critical betoken of the function $f\left( x \right)$ was a number $x = c$ so that either $f'\left( c \right) = 0$ or $f'\left( c \correct)$ doesn't exist. Nosotros have a like definition for critical points of functions of ii variables.

Definition

The point $\left( {a,b} \right)$ is a critical point (or a stationary signal) of $f\left( {10,y} \correct)$ provided i of the following is true,

$\nabla f\left( {a,b} \correct) = \vec 0$ (this is equivalent to saying that ${f_x}\left( {a,b} \correct) = 0$ and ${f_y}\left( {a,b} \right) = 0$),
${f_x}\left( {a,b} \correct)$ and/or ${f_y}\left( {a,b} \right)$ doesn't exist.

To run into the equivalence in the first part let's commencement off with $\nabla f = \vec 0$ and put in the definition of each part.

\[\begin{marshal*}\nabla f\left( {a,b} \right) & = \vec 0\\ \left\langle {{f_x}\left( {a,b} \right),{f_y}\left( {a,b} \right)} \right\rangle & = \left\langle {0,0} \right\rangle \end{align*}\]

The just manner that these 2 vectors tin can be equal is to have ${f_x}\left( {a,b} \correct) = 0$ and ${f_y}\left( {a,b} \correct) = 0$. In fact, we will use this definition of the critical bespeak more than the slope definition since it will be easier to find the critical points if we start with the partial derivative definition.

Note also that BOTH of the first guild fractional derivatives must be nada at $\left( {a,b} \right)$. If only i of the kickoff social club fractional derivatives are zero at the bespeak then the point will NOT exist a critical point.

We now accept the following fact that, at least partially, relates critical points to relative extrema.

Fact

If the bespeak $\left( {a,b} \correct)$ is a relative extrema of the function $f\left( {x,y} \right)$ and the first lodge derivatives of $f\left( {x,y} \correct)$ exist at $\left( {a,b} \correct)$ and then $\left( {a,b} \right)$ is as well a critical signal of $f\left( {x,y} \right)$ and in fact nosotros'll have $\nabla f\left( {a,b} \right) = \vec 0$.

Proof

This is a really unproblematic proof that relies on the unmarried variable version that we saw in Calculus I version, often chosen Fermat'southward Theorem.

Permit'due south showtime off past defining $g\left( x \right) = f\left( {x,b} \right)$ and suppose that $f\left( {ten,y} \right)$ has a relative extrema at $\left( {a,b} \right)$. Still, this besides means that $one thousand\left( x \right)$ also has a relative extrema (of the aforementioned kind every bit $f\left( {ten,y} \right)$) at $x = a$. By Fermat'south Theorem we then know that $g'\left( a \right) = 0$. But we too know that $thousand'\left( a \correct) = {f_x}\left( {a,b} \right)$ and so nosotros have that ${f_x}\left( {a,b} \right) = 0$.

If we now ascertain $h\left( y \right) = f\left( {a,y} \right)$ and going through exactly the same process as above we volition run across that ${f_y}\left( {a,b} \right) = 0$.

And then, putting all this together ways that $\nabla f\left( {a,b} \right) = \vec 0$ then $f\left( {x,y} \right)$ has a critical point at $\left( {a,b} \right)$.

Annotation that this does Non say that all critical points are relative extrema. Information technology only says that relative extrema volition be critical points of the office. To see this let's consider the function

\[f\left( {x,y} \right) = xy\]

The ii first order partial derivatives are,

\[{f_x}\left( {x,y} \right) = y\hspace{0.75in}{f_y}\left( {x,y} \correct) = 10\]

The but signal that volition brand both of these derivatives zero at the same time is $\left( {0,0} \right)$ so $\left( {0,0} \right)$ is a critical signal for the office. Hither is a graph of the function.

$This is a graph with the standard 3D coordinate system. The positive z-axis is straight up, the positive x-axis moves off to the right and slightly downward and positive y-axis move off the right and slightly upward. Moving into the quadrants where both x and y are positive and x and y are negative the graph flairs upwards and moving into the quadrants where and x and y have opposite signs the graph flairs downwards.$

Notation that the axes are not in the standard orientation hither so that we can meet more clearly what is happening at the origin, i.e. at $\left( {0,0} \correct)$. If we start at the origin and move into either of the quadrants where both $x$ and $y$ are the aforementioned sign the function increases. Yet, if we outset at the origin and movement into either of the quadrants where $x$ and $y$ have the opposite sign then the office decreases. In other words, no thing what region you take about the origin there will be points larger than $f\left( {0,0} \right) = 0$ and points smaller than $f\left( {0,0} \right) = 0$. Therefore, there is no way that $\left( {0,0} \correct)$ can be a relative extrema.

Critical points that exhibit this kind of behavior are called saddle points.

While we take to be careful to not misinterpret the results of this fact it is very useful in helping us to identify relative extrema. Because of this fact nosotros know that if we take all the critical points of a office then we likewise have every possible relative extrema for the part. The fact tells us that all relative extrema must be critical points and then we know that if the function does have relative extrema then they must exist in the collection of all the disquisitional points. Remember however, that it will be completely possible that at least one of the critical points won't exist a relative extrema.

So, once nosotros accept all the critical points in paw all we volition need to practise is test these points to encounter if they are relative extrema or not. To determine if a critical point is a relative extrema (and in fact to determine if information technology is a minimum or a maximum) we tin use the following fact.

Fact

Suppose that $\left( {a,b} \right)$ is a critical point of $f\left( {x,y} \right)$ and that the second order partial derivatives are continuous in some region that contains $\left( {a,b} \right)$. Adjacent ascertain,

\[D = D\left( {a,b} \right) = {f_{ten\,x}}\left( {a,b} \right){f_{y\,y}}\left( {a,b} \right) - {\left[ {{f_{10\,y}}\left( {a,b} \correct)} \correct]^2}\]

We then accept the following classifications of the critical point.

If $D > 0$ and ${f_{ten\,x}}\left( {a,b} \right) > 0$ then there is a relative minimum at $\left( {a,b} \correct)$.
If $D > 0$ and ${f_{10\,x}}\left( {a,b} \right) < 0$ then in that location is a relative maximum at $\left( {a,b} \right)$.
If $D < 0$ then the point $\left( {a,b} \right)$ is a saddle point.
If $D = 0$ then the signal $\left( {a,b} \right)$ may be a relative minimum, relative maximum or a saddle point. Other techniques would need to be used to classify the critical point.

Note that if $D > 0$ then both ${f_{\,x\,ten}}\left( {a,b} \right)$ and ${f_{\,y\,y}}\left( {a,b} \correct)$ volition accept the aforementioned sign and then in the first two cases above we could simply as easily replace ${f_{\,x\,ten}}\left( {a,b} \right)$ with ${f_{\,y\,y}}\left( {a,b} \right)$. Also note that we aren't going to be seeing any cases in this class where $D = 0$ equally these can frequently exist quite difficult to classify. Nosotros will be able to classify all the critical points that we detect.

Let's encounter a couple of examples.

Example ane Detect and classify all the critical points of $f\left( {x,y} \right) = 4 + {x^3} + {y^3} - 3xy$.

Show Solution

We first need all the first order (to find the critical points) and 2d order (to classify the disquisitional points) fractional derivatives then let's get those.

\[\brainstorm{assortment}{c}{f_x} = 3{x^2} - 3y\hspace{0.5in}{f_y} = 3{y^2} - 3x\\ {f_{x\,x}} = 6x\hspace{0.5in}{f_{y\,y}} = 6y\hspace{0.5in}{f_{x\,y}} = - iii\end{array}\]

Permit's first find the disquisitional points. Critical points volition be solutions to the arrangement of equations,

\[\begin{align*}{f_x} & = 3{ten^2} - 3y = 0\\ {f_y} & = iii{y^two} - 3x = 0\end{align*}\]

This is a non-linear system of equations and these can, on occasion, be difficult to solve. Nonetheless, in this case it's not likewise bad. Nosotros tin can solve the beginning equation for $y$ as follows,

\[three{x^two} - 3y = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}y = {x^ii}\]

Plugging this into the second equation gives,

\[iii{\left( {{x^2}} \right)^2} - 3x = 3x\left( {{ten^three} - 1} \correct) = 0\]

From this we can see that we must have $x = 0$ or $10 = 1$. At present use the fact that $y = {x^2}$ to get the critical points.

\[\begin{align*}x = 0: & \hspace{0.25in}y = {0^2} = 0\hspace{0.1in} & \Rightarrow & \hspace{0.5in}\left( {0,0} \right)\\ x = ane: & \hspace{0.25in}y = {one^2} = 1\hspace{0.1in} & \Rightarrow & \hspace{0.5in}\left( {1,1} \right)\cease{align*}\]

So, we get two critical points. All we demand to practise now is allocate them. To exercise this we will demand $D$. Here is the general formula for $D$.

\[\begin{align*}D\left( {x,y} \right) & = {f_{x\,x}}\left( {x,y} \correct){f_{y\,y}}\left( {x,y} \right) - {\left[ {{f_{x\,y}}\left( {ten,y} \right)} \right]^2}\\ & = \left( {6x} \right)\left( {6y} \right) - {\left( { - 3} \right)^2}\\ & = 36xy - ix\end{align*}\]

To classify the critical points all that we need to exercise is plug in the critical points and apply the fact above to allocate them.

$\left( {0,0} \right)$ : \[D = D\left( {0,0} \correct) = - nine < 0\]

Then, for $\left( {0,0} \correct)$ $D$ is negative and and so this must be a saddle point.

$\left( {1,i} \right)$ : \[D = D\left( {ane,1} \right) = 36 - 9 = 27 > 0\hspace{0.5in}{f_{ten\,ten}}\left( {1,1} \right) = 6 > 0\]

For $\left( {1,one} \correct)$ $D$ is positive and ${f_{x\,ten}}$ is positive and so we must take a relative minimum.

For the sake of abyss here is a graph of this role.

$This is a graph with the standard 3D coordinate system. The positive z-axis is straight up, the positive x-axis moves off to the right and slightly downward and positive y-axis move off the right and slightly upward. Most of this graph is a vertical plane that is at an approximately 45 degree angle with the x-axis. On the right side of this plane a vaguely cup shaped object protrudes into a valley that occurs over (1,1) in the xy-plane.$ $This is a boxed 3D coordinate system. The z-axis is right vertical edge of the box, the x-axis is the top back edge of the box and the y-axis is the bottom left edge of the box. Most of this graph is a vertical plane that is at an approximately 45 degree angle with the x-axis. On the right side of this plane a vaguely cup shaped object protrudes into a valley that occurs over (1,1) in the xy-plane.$

Notice that in order to become a meliorate visual we used a somewhat nonstandard orientation. We can come across that there is a relative minimum at $\left( {1,1} \right)$ and (hopefully) it's clear that at $\left( {0,0} \right)$ we do get a saddle point.

Case 2 Find and allocate all the critical points for $f\left( {x,y} \right) = 3{x^2}y + {y^iii} - 3{x^2} - 3{y^2} + 2$

Show Solution

As with the first example we will kickoff need to get all the first and 2nd order derivatives.

\[\begin{array}{c}{f_x} = 6xy - 6x\hspace{0.5in}{f_y} = 3{x^ii} + 3{y^2} - 6y\\ {f_{x\,10}} = 6y - half dozen\hspace{0.5in}{f_{y\,y}} = 6y - 6\hspace{0.75in}{f_{x\,y}} = 6x\end{array}\]

Nosotros'll start need the disquisitional points. The equations that we'll demand to solve this time are,

\[\brainstorm{align*}6xy - 6x & = 0\\ three{x^ii} + three{y^2} - 6y & = 0\end{align*}\]

These equations are a little trickier to solve than the first set, but in one case you see what to do they really aren't terribly bad.

Get-go, let's notice that we can factor out a 6$x$ from the first equation to go,

\[6x\left( {y - i} \right) = 0\]

Then, nosotros can run across that the starting time equation will be nix if $x = 0$ or $y = 1$. Be careful to not just cancel the $x$ from both sides. If nosotros had washed that we would have missed $x = 0$.

To find the critical points we can plug these (individually) into the second equation and solve for the remaining variable.

$x = 0$ : \[iii{y^2} - 6y = 3y\left( {y - two} \correct) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}y = 0,\,\,y = 2\] $y = 1$ : \[3{x^2} - 3 = 3\left( {{x^2} - i} \right) = 0\hspace{0.5in} \Rightarrow \hspace{0.25in}x = - 1,\,\,x = 1\]

So, if $ten = 0$ we accept the following critical points,

\[\left( {0,0} \correct)\hspace{0.25in}\left( {0,2} \right)\]

and if $y = i$ the disquisitional points are,

\[\left( {one,one} \correct)\hspace{0.25in}\left( { - ane,i} \right)\]

Now all we need to exercise is classify the disquisitional points. To do this we'll need the general formula for $D$.

\[D\left( {x,y} \right) = \left( {6y - half-dozen} \correct)\left( {6y - 6} \right) - {\left( {6x} \correct)^two} = {\left( {6y - 6} \right)^2} - 36{x^2}\] $\left( {0,0} \right)$ : \[D = D\left( {0,0} \right) = 36 > 0\hspace{0.5in}{f_{x\,ten}}\left( {0,0} \right) = - 6 < 0\] $\left( {0,ii} \right)$ : \[D = D\left( {0,2} \right) = 36 > 0\hspace{0.5in}{f_{ten\,ten}}\left( {0,2} \correct) = half dozen > 0\] $\left( {1,1} \correct)$ : \[D = D\left( {i,1} \correct) = - 36 < 0\] $\left( { - i,1} \right)$ : \[D = D\left( { - 1,1} \right) = - 36 < 0\]

So, information technology looks like we have the following nomenclature of each of these critical points.

\[\begin{align*} & \left( {0,0} \right) & & :\hspace{0.5in}{\mbox{Relative Maximum}}\\ & \left( {0,2} \correct) & & : \hspace{0.5in} {\mbox{Relative Minimum}}\\ & \left( {i,one} \right) & & :\hspace{0.5in}{\mbox{Saddle Point}}\\ & \left( { - 1,1} \right) & & :\hspace{0.5in}{\mbox{Saddle Point}}\end{marshal*}\]

Hither is a graph of the surface for the sake of completeness.

$This is a graph with the standard 3D coordinate system. The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis move off the right and slightly downward. This object looks almost like two cup shaped objects that have been glued together. On the right side is a cup shaped object that opens upwards with a valley at (0,2) in the xy-plane. On the left side is a cup shaped object that opens downwards with a peak at (0,0) in the xy-plane. The peak of this cup shaped object is higher than the valley of the first cup shaped object.$ $This is a boxed 3D coordinate system. The z-axis is left vertical edge of the box, the x-axis is the top left edge of the box and the y-axis is the bottom front edge of the box. This object looks almost like two cup shaped objects that have been glued together. On the right side is a cup shaped object that opens upwards with a valley at (0,2) in the xy-plane. On the left side is a cup shaped object that opens downwards with a peak at (0,0) in the xy-plane. The peak of this cup shaped object is higher than the valley of the first cup shaped object.$

Let'south do one more than instance that is a petty unlike from the kickoff ii.

Example 3 Determine the point on the aeroplane $4x - 2y + z = 1$ that is closest to the point $\left( { - 2, - 1,5} \right)$.

Show Solution

Note that we are NOT request for the critical points of the plane. In club to exercise this case we are going to demand to first come up with the equation that nosotros are going to have to work with.

Outset, allow'southward suppose that $\left( {ten,y,z} \correct)$ is whatever bespeak on the plane. The distance between this signal and the betoken in question, $\left( { - 2, - 1,five} \correct)$, is given by the formula,

\[d = \sqrt {{{\left( {x + 2} \correct)}^2} + {{\left( {y + 1} \right)}^2} + {{\left( {z - 5} \correct)}^two}} \]

What we are then asked to observe is the minimum value of this equation. The point $\left( {x,y,z} \right)$ that gives the minimum value of this equation will be the bespeak on the plane that is closest to $\left( { - 2, - 1,5} \right)$.

There are a couple of problems with this equation. Commencement, it is a function of $10$, $y$ and $z$ and we can only bargain with functions of $10$ and $y$ at this point. However, this is easy to fix. Nosotros tin can solve the equation of the plane to see that,

\[z = i - 4x + 2y\]

Plugging this into the distance formula gives,

\[\begin{marshal*}d & = \sqrt {{{\left( {x + ii} \right)}^2} + {{\left( {y + i} \right)}^two} + {{\left( {1 - 4x + 2y - 5} \correct)}^two}} \\ & = \sqrt {{{\left( {x + ii} \right)}^2} + {{\left( {y + one} \right)}^ii} + {{\left( { - 4 - 4x + 2y} \right)}^2}} \finish{align*}\]

At present, the adjacent event is that there is a square root in this formula and we know that we're going to exist differentiating this eventually. And so, in guild to brand our life a little easier allow's notice that finding the minimum value of $d$ volition be equivalent to finding the minimum value of ${d^2}$.

And so, let'south instead find the minimum value of

\[f\left( {x,y} \right) = {d^2} = {\left( {x + 2} \correct)^2} + {\left( {y + ane} \correct)^two} + {\left( { - 4 - 4x + 2y} \right)^2}\]

Now, we demand to be a trivial conscientious here. We are being asked to find the closest indicate on the plane to $\left( { - 2, - 1,5} \right)$ and that is non actually the same thing as what nosotros've been doing in this section. In this section we've been finding and classifying disquisitional points equally relative minimums or maximums and what we are really request is to detect the smallest value the function will take, or the absolute minimum. Hopefully, it does brand sense from a physical standpoint that in that location will be a closest betoken on the plane to $\left( { - two, - 1,five} \right)$. This indicate should also exist a relative minimum in add-on to being an absolute minimum.

So, let's go through the procedure from the offset and second instance and encounter what nosotros get as far as relative minimums go. If we but get a single relative minimum and so nosotros volition exist done since that point volition also demand to exist the accented minimum of the function and hence the point on the plane that is closest to $\left( { - ii, - 1,5} \right)$.

We'll need the derivatives beginning.

\[\brainstorm{align*}{f_x} & = 2\left( {ten + 2} \right) + 2\left( { - iv} \correct)\left( { - four - 4x + 2y} \right) = 36 + 34x - 16y\\ {f_y} & = 2\left( {y + 1} \right) + ii\left( ii \right)\left( { - 4 - 4x + 2y} \right) = - 14 - 16x + 10y\\ {f_{x\,x}} & = 34\\ {f_{y\,y}} & = ten\\ {f_{ten\,y}} & = - sixteen\end{marshal*}\]

Now, before we get into finding the critical point let's compute $D$ quickly.

\[D = 34\left( {10} \correct) - {\left( { - xvi} \right)^2} = 84 > 0\]

So, in this case $D$ volition ever exist positive and besides notice that ${f_{x\,ten}} = 34 > 0$ is always positive and and then any critical points that nosotros get will be guaranteed to be relative minimums.

Now let's observe the critical point(southward). This will mean solving the system.

\[\brainstorm{align*}36 + 34x - 16y & = 0\\ - 14 - 16x + 10y & = 0\stop{align*}\]

To do this nosotros can solve the kickoff equation for $x$.

\[x = \frac{1}{{34}}\left( {16y - 36} \right) = \frac{1}{{17}}\left( {8y - 18} \right)\]

Now, plug this into the second equation and solve for $y$.

\[ - 14 - \frac{{16}}{{17}}\left( {8y - 18} \correct) + 10y = 0\hspace{0.5in} \Rightarrow \hspace{0.5in}y = - \frac{{25}}{{21}}\]

Dorsum substituting this into the equation for $10$ gives $x = - \frac{{34}}{{21}}$.

And so, information technology looks like we get a single critical point : $\left( { - \frac{{34}}{{21}}, - \frac{{25}}{{21}}} \right)$. As well, since nosotros know this volition be a relative minimum and it is the only critical point nosotros know that this is too the $ten$ and $y$ coordinates of the bespeak on the airplane that we're after. We can find the $z$ coordinate past plugging into the equation of the airplane as follows,

\[z = 1 - 4\left( { - \frac{{34}}{{21}}} \correct) + 2\left( { - \frac{{25}}{{21}}} \right) = \frac{{107}}{{21}}\]

So, the bespeak on the plane that is closest to $\left( { - 2, - 1,five} \right)$ is $\left( { - \frac{{34}}{{21}}, - \frac{{25}}{{21}},\frac{{107}}{{21}}} \right)$.