Wize University Calculus 3 Textbook > Applications of Partial Derivatives

Maximum and Minimum Points - Finding Critical Points in Level Curves

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Maximums, Minimums, and Critical Points

A function has a relative minimum at the point (a, b) if f(x, y)≥f(a, b)f(x,~y)\geq f(a,~b)f(x, y)≥f(a, b) for all points (x, y) in some region around (a, b).
A function has a relative maximum at the point (a, b) if f(x, y)≤f(a, b) ]f(x,~y)\leq f(a,~b)~]f(x, y)≤f(a, b) ] for all points (x, y) in some region around (a, b).
The point (a, b) is a critical point of f(x, y), provided one is true:
∇f(a, b)=0⃗\nabla f(a,~b)=\vec{0}∇f(a, b)=0
fx(a, b) f_x(a,~b)~fx​(a, b) and/or fy(a, b) f_y(a,~b)~fy​(a, b) does not exist.
Watch Out!
Not all critical points are relative extrema but all relative extrema are critical points

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Classifying Critical Points
Suppose that (a, b) (a,~b)~(a, b) is a critical point of f(x, y)f(x,~y)f(x, y) and that the 2nd^{\text{nd}}nd order partial derivatives are continuous in some region that contains (a, b)(a,~b)(a, b).

Then, the following is used to classify critical points:
D=D(a, b)=fxx(a, b)fyy(a, b)−[fxy(a, b)]2\boxed{D=D(a,~b)=f_{xx}(a,~b)f_{yy}(a,~b)-\Big[f_{xy}(a,~b)\Big]^2}D=D(a, b)=fxx​(a, b)fyy​(a, b)−[fxy​(a, b)]2​
If D>0 & fxx(a, b)>0,D>0~\&~f_{xx}(a,~b)>0,D>0 & fxx​(a, b)>0, there is a relative minimum at (a, b).
If D>0 & fxx(a, b)<0,D>0~\&~f_{xx}(a,~b)<0,D>0 & fxx​(a, b)<0, there is a relative maximum at (a, b).
If D<0,D<0,D<0, then (a, b) is a saddle point.
If D=0,D=0,D=0, then (a, b) could be a relative maximum, relative minimum, or a saddle point but other techniques will need to be used to classify (a, b). 
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Level Curves & Gradient Vectors
Critical points can also be classified in the level curves using the gradient vectors.
Gradient vectors are always orthogonal to level curves.
Gradients vectors are pointing toward the relative maximum in all directions.
Gradient vectors are pointing away from relative minimum in all directions.
For saddle points, gradient vectors are pointing toward the point in some directions and away from the point in other directions.

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Example
Find and classify all critical points of f(x, y)=x3+y3−3xy.f(x,~y)=x^3+y^3-3xy.f(x, y)=x3+y3−3xy.

Step 1: Find all the first and second order partial derivatives
fx=3x2−3yfy=3y2−3xfxx=6xfyy=6yfxy=−3\begin{array}{c|c|c}
f_x=3x^2-3y & f_y=3y^2-3x & \\f_{xx}=6x & f_{yy}=6y & f_{xy}=-3
\end{array}fx​=3x2−3yfxx​=6x​fy​=3y2−3xfyy​=6y​fxy​=−3​
Step 2: Find the critical points
fx=0=3x2−3y   →   y=x2fy=0=3y2−3x                     Solve by substitution3(x2)2−3x=03x(x3−1)=0x=0, 1∴ (0, 0) and (1, 1) are critical pointsf_x=0=3x^2-3y~~~{\color{red}\rightarrow}~~~y=x^2\newline
f_y=0=3y^2-3x~~~~~~~~~~~~~~~~~~~~~\newline{}
{\color{magenta}\text{\footnotesize{Solve by substitution}}}\newline{}\newline{}
3(x^2)^2-3x=0\newline
3x(x^3-1)=0\newline{}
x=0,~1\newline{}\newline{}
\therefore~(0,~0)~\text{and}~(1,~1)~\text{are critical points}
fx​=0=3x2−3y   →   y=x2fy​=0=3y2−3x                     Solve by substitution3(x2)2−3x=03x(x3−1)=0x=0, 1∴ (0, 0) and (1, 1) are critical points
Step 3: Classify the critical points
D(a, b)=fxx(a, b)fyy(a, b)−[fxy(a, b)]2=(6x)(6y)−(−3)2=36xy−9D(a,~b)=f_{xx}(a,~b)f_{yy}(a,~b)-\Big[f_{xy}(a,~b)\Big]^2\newline
=(6x)(6y)-(-3)^2\newline
=36xy-9D(a, b)=fxx​(a, b)fyy​(a, b)−[fxy​(a, b)]2=(6x)(6y)−(−3)2=36xy−9

At (0, 0)‾:                        At (1, 1)‾:D(0, 0)=−9<0    D(1, 1)=36−9>0fxx(1, 1)=6(1)>0fyy(1, 1)=6(1)>0∴ (0, 0) is a saddle point∴ (1, 1) is a relative minimum\begin{array}{c|c}
\text{\color{magenta}\underline{At (0,~0)}}:~~~~~~~~~~~~~~~~~~~~~~~~ & \text{\color{magenta}\underline{At (1,~1)}}:\\\\
D(0,~0)=-9<0~~~~& D(1,~1)=36-9>0\\\\
&f_{xx}(1,~1)=6(1)>0\\
&f_{yy}(1,~1)=6(1)>0\\\\
\therefore ~(0,~0) ~\text{is a saddle point} & \therefore ~ (1,~1)~\text{is a relative minimum}
\end{array}At (0, 0)​:                        D(0, 0)=−9<0    ∴ (0, 0) is a saddle point​At (1, 1)​:D(1, 1)=36−9>0fxx​(1, 1)=6(1)>0fyy​(1, 1)=6(1)>0∴ (1, 1) is a relative minimum​

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Example
Determine the point on the plane 4x−2y+z=1 4x-2y+z=1~4x−2y+z=1 closest to (−2, −1, 5).(-2,~-1,~5).(−2, −1, 5).

Let (x, y, z) (x,~y,~z)~(x, y, z) be the point on the plane closest to (−2, −1, 5). (-2,~-1,~5).~(−2, −1, 5). The distance between them is:
d2=(x+2)2+(y+1)2+(z−5)2d^2=(x+2)^2+(y+1)^2+(z-5)^2d2=(x+2)2+(y+1)2+(z−5)2
The equation of the plane can be written as:
z=−4x+2y+1z=-4x+2y+1z=−4x+2y+1
Substitute the equation of the plane into the formula we have for distance:
d2=(x+2)2+(y+1)2+(−4x+2y+1−5)2d2=(x+2)2+(y+1)2+(−4x+2y−4)2d^2=(x+2)^2+(y+1)^2+(-4x+2y+1-5)^2\newline
d^2=(x+2)^2+(y+1)^2+(-4x+2y-4)^2d2=(x+2)2+(y+1)2+(−4x+2y+1−5)2d2=(x+2)2+(y+1)2+(−4x+2y−4)2

Watch Out!
We want the point on the plane closest to (-2, -1, 5), which is not the same thing as a critical point. We will use the critical point to find the point on the plane that we are looking for. 

Find the partial derivatives:
fx=34x−16y+36fy=−16x+10y−14fxx=34fyy=10fxy=−16\begin{array}{c|c|c}
f_x=34x-16y+36 & f_y=-16x+10y-14&\\
f_{xx}=34 & f_{yy}=10 & f_{xy}=-16
\end{array}
fx​=34x−16y+36fxx​=34​fy​=−16x+10y−14fyy​=10​fxy​=−16​

Notice that fxx>0, fyy>0, and D=(34)(10)−(−16)2=84>0 f_{xx}>0,~f_{yy}>0,~\text{and}~D=(34)(10)-(-16)^2=84>0~fxx​>0, fyy​>0, and D=(34)(10)−(−16)2=84>0 always, which means that any critical point we get will be a relative minimum ⌣¨\ddot\smile⌣¨

Find the critical points:
fx=0=36+34x−16y    →    (17x−8y=−18)×5fy=0=−14−16x+10y    →    (−8x+5y=7)×885x−40y=−90−(−64+40y=56)‾21x=−34x=−3421∴ y=−2521f_x=0=36+34x-16y~~~~{\color{red}\rightarrow}~~~~(17x-8y=-18)\times5\newline
f_y=0=-14-16x+10y~~~~{\color{red}\rightarrow~~~~}(-8x+5y=7)\times8\newline{}\newline{}
85x-40y=-90\newline
\underline{-(-64+40y=56)}\newline{}
21x=-34\newline{}\newline{}x=-\frac{34}{21}\newline{}\newline{}\therefore~y=\frac{-25}{21}fx​=0=36+34x−16y    →    (17x−8y=−18)×5fy​=0=−14−16x+10y    →    (−8x+5y=7)×885x−40y=−90−(−64+40y=56)​21x=−34x=−2134​∴ y=21−25​

Ok, so where is our critical point (−3421, −2521)? \Big(-\frac{34}{21},~-\frac{25}{21}\Big)?~(−2134​, −2125​)? We can substitute the critical point into the equation of the plane find out:
z=1−4x+2yz=1−4(−3421)+2(−2521)]z=10721z=1-4x+2y\newline{}\newline{}z=1-4\Big(-\frac{34}{21}\Big)+2\Big(-\frac{25}{21}\Big)]\newline{}\newline{}
z=\frac{107}{21}z=1−4x+2yz=1−4(−2134​)+2(−2125​)]z=21107​
So the point closest is 
(−3421, −2521, 10721)\Big(-\frac{34}{21},~-\frac{25}{21},~\frac{107}{21}\Big)(−2134​, −2125​, 21107​)

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Practice
Find and classify all the critical points for f(x, y)=(y−2)x2−y2f(x,~y)=(y-2)x^2-y^2f(x, y)=(y−2)x2−y2.

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Practice
Determine the point on the plane 3x+10y+z=9 3x+10y+z=9~3x+10y+z=9 closest to (1, 1, 5)(1,~1,~5)(1, 1, 5).

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Wize University Calculus 3 Textbook > Applications of Partial Derivatives

Maximum and Minimum Points - Finding Critical Points in Level Curves

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