Wize University Calculus 3 Textbook > Applications of Partial Derivatives

Absolute Maximum and Minimum Values

Maximum and Minimum Points - Finding Critical Points in Level CurvesPrevious Section

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Absolute Max and Min
In order to optimize a function, we need the functions absolute maximum and\or minimum. 

In order to find the absolute maximum/minimum values of the continuous function f(x,y)f(x,y)f(x,y) over a closed boundary, the following steps should be taken:

Step 1 
Find the critical points of f(x,y)\colorThree{f(x,y)}f(x,y) over D\colorThree{D}D and determine the function value at each of these points

Step 2 
Find the values of f(x,y)\colorThree{f(x,y)}f(x,y) at the boundaries of D\colorThree{D}D and all critical points

Step 3 
The largest value of the function found in the previous step is the absolute maximum value, while the smallest value found is the absolute minimum value.
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The Extreme Value Theorem
The Extreme Value Theorem states that if f(x, y)f(x,~y)f(x, y) is continuous in some closed, bounded set D D~D in R2 \mathbb{R}^2~R2 , then there are points in D(x1, y1) D(x_1,~y_1)~D(x1​, y1​) and D(x2, y2) D(x_2,~y_2)~D(x2​, y2​) so that f(x1, y1) f(x_1,~y_1)~f(x1​, y1​) is the absolute maximum and f(x2, y2) f(x_2,~y_2)~f(x2​, y2​) is the absolute minimum of the function D.D.D.

Wize Tip
The Extreme Value Theorem just tells us about the existence of an absolute max/min but it does NOT tell us where it will occur.

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Example
Find the absolute maximum and absolute minimum of f(x,y)=x2−2xy+yf\left(x,y\right)=x^2-2xy+yf(x,y)=x2−2xy+y over
D={(x,y)∣ ∣x∣≤1, ∣y∣≤1}D=\left\{\left(x,y\right)\left|\ \left|x\right|\le1,\ \left|y\right|\le1\right.\right\}D={(x,y)∣ ∣x∣≤1, ∣y∣≤1}

f(x,y)=x2−2xy+yD={(x,y)∣  ∣x∣≤1,∣y∣≤1}f(x,y)=x^2-2xy+y\quad D=\{(x,y)|\;|x|\leq 1, |y|\leq 1\}f(x,y)=x2−2xy+yD={(x,y)∣∣x∣≤1,∣y∣≤1}

Let's find critical points
fx=2x−2y=0⇒x=yf_x=2x-2y=0\Rightarrow x=yfx​=2x−2y=0⇒x=y
fy=−2x+1=0⇒x=12f_y=-2x+1=0\Rightarrow x=\frac{1}{2}fy​=−2x+1=0⇒x=21​
⇒(12,12)\Rightarrow \left(\frac{1}{2},\frac{1}{2}\right)⇒(21​,21​) critical point it's inside domain DDD
f(12,12)=14−2⋅12⋅12+12=14 f\left(\frac{1}{2},\frac{1}{2}\right)=\frac{1}{4}-2\cdot\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}=\frac{1}{4}f(21​,21​)=41​−2⋅21​⋅21​+21​=41​  (eq. 1)

Now we have to look at max & min points on each boundary line!

Right: x=1,−1≤y≤1x=1, \quad -1\leq y\leq 1x=1,−1≤y≤1

⇒f(1,y)=g1(y)=1−2y+y=1−y\Rightarrow f(1,y)=g_1(y)=1-2y+y=1-y⇒f(1,y)=g1​(y)=1−2y+y=1−y
g1max(y=−1)=f(1,−1)=2g_1^{\mathrm{max}}(y=-1)=f(1,-1)=2g1max​(y=−1)=f(1,−1)=2 (eq. 2)
g1min(y=1)=f(1,1)=0g_1^{\mathrm{min}}(y=1)=f(1,1)=0g1min​(y=1)=f(1,1)=0 (eq. 3)

Left: x=−1,−1≤y≤1x=-1, \quad -1\leq y\leq 1x=−1,−1≤y≤1

⇒f(−1,y)=g2(y)=1+2y+y=1+3y\Rightarrow f(-1,y)=g_2(y)=1+2y+y=1+3y⇒f(−1,y)=g2​(y)=1+2y+y=1+3y
g2max(y=1)=f(−1,1)=4g_2^{\mathrm{max}}(y=1)=f(-1,1)=4g2max​(y=1)=f(−1,1)=4 (eq. 4)
g2min(y=−1)=f(−1,−1)=−2g_2^{\mathrm{min}}(y=-1)=f(-1,-1)=-2g2min​(y=−1)=f(−1,−1)=−2 (eq. 5)

Up: −1≤x≤1,y=1-1\leq x\leq 1, \quad y=1−1≤x≤1,y=1

⇒f(x,1)=g3(x)=x2−2x+1=(x−1)2\Rightarrow f(x,1)=g_3(x)=x^2-2x+1=(x-1)^2⇒f(x,1)=g3​(x)=x2−2x+1=(x−1)2
g3max(x=−1)=f(−1,1)=4g_3^{\mathrm{max}}(x=-1)=f(-1,1)=4g3max​(x=−1)=f(−1,1)=4
g3min(x=1)=f(1,1)=0g_3^{\mathrm{min}}(x=1)=f(1,1)=0g3min​(x=1)=f(1,1)=0

Down: −1≤x≤1,y=−1-1\leq x\leq 1, \quad y=-1−1≤x≤1,y=−1

⇒f(x,−1)=g4(x)=x2+2x−1\Rightarrow f(x,-1)=g_4(x)=x^2+2x-1⇒f(x,−1)=g4​(x)=x2+2x−1
g4max(x=1)=f(1,−1)=2g_4^{\mathrm{max}}(x=1)=f(1,-1)=2g4max​(x=1)=f(1,−1)=2
g4min(x=−1)=f(−1,−1)=−2g_4^{\mathrm{min}}(x=-1)=f(-1,-1)=-2g4min​(x=−1)=f(−1,−1)=−2

from equation (1) to (5):
absolute max=max⁡{14,0,2,−2,4}=4at(−1,1)\mathrm{absolute\ max}=\max\left\lbrace \frac{1}{4},0,2,-2,4 \right\rbrace=4 \quad \mathrm{at} \quad (-1,1)absolute max=max{41​,0,2,−2,4}=4at(−1,1)
absolute min=min⁡{14,0,2,−2,4}=−2at(−1,−1)\mathrm{absolute\ min}=\min\left\lbrace \frac{1}{4},0,2,-2,4\right\rbrace=-2 \quad \mathrm{at} \quad (-1,-1)absolute min=min{41​,0,2,−2,4}=−2at(−1,−1)

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Example
Find the absolute maximum and minimum of f(x, y)=(9x2−1)(1+4y) f(x,~y)=(9x^2-1)(1+4y)~f(x, y)=(9x2−1)(1+4y) on the domain −2≤x≤3 and −1≤y≤4-2\leq{x}\leq{3}~\text{and}~-1\leq{y}\leq{4}−2≤x≤3 and −1≤y≤4

The domain is given by the following:
  
So, let's first find the critical points:
fx=18x(1+4y)fy=4(9x2−1)    →    x=±13f_x=18x(1+4y)\newline{}
f_y=4(9x^2-1)~~~~{\color{red}\rightarrow}~~~~x=\pm\frac{1}{3}fx​=18x(1+4y)fy​=4(9x2−1)    →    x=±31​
Let us test to see if these critical points lie in the domain:
f(13, y)=0=18(13)(1+4y)f(−13, y)=0=18(−13)(1+4y)y=−14y=−14\begin{array}{c|c}
f(\frac{1}{3},~y)=0=18(\frac{1}{3})(1+4y) & f(-\frac{1}{3},~y)=0=18(-\frac{1}{3})(1+4y)\\\\y=-\frac{1}{4} & y=-\frac{1}{4}
\end{array}f(31​, y)=0=18(31​)(1+4y)y=−41​​f(−31​, y)=0=18(−31​)(1+4y)y=−41​​

The critical points, (13, −14) and (−13, −14), \Big(\frac{1}{3},~-\frac{1}{4}\Big)~\text{and}~\Big(-\frac{1}{3},~-\frac{1}{4}\Big),~(31​, −41​) and (−31​, −41​), are in the domain.

Now, we need to check the boundaries:
y=4,  −2≤x≤3y=−1,  −2≤x≤3x=−2,  −1≤y≤4x=3,  −1≤y≤4y=4,~~-2\leq{x}\leq{3}\newline{}
y=-1,~~-2\leq{x}\leq{3}\newline{}
x=-2,~~-1\leq{y}\leq{4}\newline{}
x=3,~~-1\leq{y}\leq{4}\newline{}y=4,  −2≤x≤3y=−1,  −2≤x≤3x=−2,  −1≤y≤4x=3,  −1≤y≤4
1. y=4,  −2≤x≤3‾:\underline{y=4,~~-2\leq{x}\leq{3}}:y=4,  −2≤x≤3​:
g(x)=f(x, 4)=17(9x2−1)g′(x)=fx(x, 4)=306x=0∴ x=0g(x)=f(x,~4)=17(9x^2-1)\newline{}
g'(x)=f_x(x,~4)=306x=0\newline{}
\therefore~x=0g(x)=f(x, 4)=17(9x2−1)g′(x)=fx​(x, 4)=306x=0∴ x=0
     The point (0, 4) is in the domain. Thus, 
f(0, 4)=−17f(0,~4)=-17f(0, 4)=−17
     We also test the vertex points:
f(−2, 4)=595f(3, 4)=1360f(-2,~4)=595\newline{}
f(3,~4)=1360f(−2, 4)=595f(3, 4)=1360

2. y=−1,  −2≤x≤3‾:\underline{y=-1,~~-2\leq{x}\leq{3}}:y=−1,  −2≤x≤3​:
g(x)=f(x, −1)=−3(9x2−1)g′(x)=fx(x, −1)=−54x=0∴ x=0g(x)=f(x,~-1)=-3(9x^2-1)\newline{}
g'(x)=f_x(x,~-1)=-54x=0\newline{}
\therefore~x=0g(x)=f(x, −1)=−3(9x2−1)g′(x)=fx​(x, −1)=−54x=0∴ x=0
    The point (0, -1) is in the domain. Thus, 
f(0, −1)=3f(0,~-1)=3f(0, −1)=3
    We also text the vertex points:
f(−2, −1)=−105f(3, −1)=−240f(-2,~-1)=-105\newline{}
f(3,~-1)=-240f(−2, −1)=−105f(3, −1)=−240

3. x=−2,  −1≤y≤4‾:\underline{x=-2,~~-1\leq{y}\leq{4}}:x=−2,  −1≤y≤4​:
h(y)=f(−2, y)=35(1+4y)h′(y)=fy(−2, y)=140No critical points hereh(y)=f(-2,~y)=35(1+4y)\newline{}
h'(y)=f_y(-2,~y)=140\newline{}
{\color{magenta}\text{\footnotesize{No critical points here}}}h(y)=f(−2, y)=35(1+4y)h′(y)=fy​(−2, y)=140No critical points here

4. x=3,  −1≤y≤4‾:\underline{x=3,~~-1\leq{y}\leq{4}}:x=3,  −1≤y≤4​:
h(y)=f(3, y)=80(1+4y)h′(y)=fy(3, y)=320No critical points hereh(y)=f(3,~y)=80(1+4y)\newline{}
h'(y)=f_y(3,~y)=320\newline{}
{\color{magenta}\text{\footnotesize{No critical points here}}}h(y)=f(3, y)=80(1+4y)h′(y)=fy​(3, y)=320No critical points here

Thus, the absolute maximum is f(3, 4)=1360 f(3,~4)=1360~f(3, 4)=1360 and the absolute minimum is f(3, −1)=−240f(3,~-1)=-240f(3, −1)=−240

Practice
a) Find the complete list of points where the absolute maximum/minimum of f(x, y)=3x2+2xy−10y2 f(x,~y)=3x^2+2xy-10y^2~f(x, y)=3x2+2xy−10y2 on the triangle with vertices (2, 2), (2, −2), (5,−2) (2,~2),~(2,~-2),~(5,-2)~(2, 2), (2, −2), (5,−2) could occur.


b) Find the absolute maximum and minimum of f(x, y)=3x2+2xy−10y2 f(x,~y)=3x^2+2xy-10y^2~f(x, y)=3x2+2xy−10y2 on the triangle with vertices (2, 2), (2, −2), (5,−2).(2,~2),~(2,~-2),~(5,-2).(2, 2), (2, −2), (5,−2).

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a) Find the complete list of points where the absolute maximum/minimum of f(x, y)=3x2+2xy−10y2 f(x,~y)=3x^2+2xy-10y^2~f(x, y)=3x2+2xy−10y2 on the triangle with vertices (2, 2), (2, −2), (5,−2) (2,~2),~(2,~-2),~(5,-2)~(2, 2), (2, −2), (5,−2) could occur.

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Wize University Calculus 3 Textbook > Applications of Partial Derivatives

Absolute Maximum and Minimum Values

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Absolute Max and Min

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