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Wize University Calculus 2 Textbook > Multivariable Functions

Max & Min

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Local Max & Min

  • A function f(x,y)f\left(x,y\right) has a local max at (a,b)\left(a,b\right) if f(x,y)≤f(a,b)f\left(x,y\right)\le f\left(a,b\right) near the point (a,b)\left(a,b\right)
  • A function f(x,y)f\left(x,y_{ }\right) has a local min at (a,b)\left(a,b\right) if f(x,y)≥f(a,b)f\left(x,y\right)\ge f\left(a,b\right) near the point (a,b)\left(a,b\right)
  • If ff has a local max or min at the point (a,b)\left(a,b\right), then there is a critical point at (a,b)\left(a,b\right) and fx(a,b)=fy(a,b)=0f_x\left(a,b\right)=f_y\left(a,b\right)=0
Second Derivative Test
Suppose that ff has a critical point at (a,b)\left(a,b\right).
Let D(a,b)=fxx(a,b)fyy(a,b)−[fxy(a,b)]2D\left(a,b\right)=f_{xx}\left(a,b\right)f_{yy}\left(a,b\right)-\left[f_{xy}\left(a,b\right)\right]^2.
DfxxConclusion++f(a,b) is a local min+−f(a,b) is a local max−f(a,b) is a saddle point*It’s a local min or max depending onthe path you take towards the point\begin{array}{|c|c|c|} \hline \bold{D}&\bold{f_{xx}}&\bold{\text{Conclusion}}\\\hline +&+&f(a,b)\text{ is a \textcolor{orange}{local min}}\\\hline +&-&f(a,b)\text{ is a \textcolor{orange}{local max}}\\\hline -&&\begin{array}{c}f(a,b)\text{ is a \textcolor{orange}{saddle point}}\\\text{*It's a local min or max depending on}\\ \text{the path you take towards the point}\end{array}\\\hline \end{array}

Example
Identify all local max, min, and saddle points of the function f(x,y)=x2−4xy+y2f\left(x,y\right)=x^2-4xy+y^2
Critical points occur when the first partials equal 0:
  • fx=2x−4y=0f_x=2x-4y=0
  • fy=−4x+2y=0 f_y=-4x+2y=0\
Solving these 2 equations, we see that the only critical point is (0,0)\left(0,0\right)

Second partials:
  • fxx=2f_{xx}=2
  • fyy=2f_{yy}=2
  • fxy=−4f_{xy}=-4
D=fxx(0,0)fyy(0,0)−[fxy(0,0)]2=(2)(2)−(−4)2=−12D=f_{xx}\left(0,0\right)f_{yy}\left(0,0\right)-\left[f_{xy}\left(0,0\right)\right]^2=(2)(2)-(-4)^2=-12
Since DD is negative at this critical point, we know that (0,0)\left(0,0\right) is a saddle point.
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Absolute Max & Min

If ff is continuous on a closed and bounded domain, then ff attains an absolute max and absolute min values at some points on this domain.

Steps to finding the absolute max & min
  1. Find all critical points (x,y)(x,y) of ff on the domain, and find their corresponding ff values
  2. Find the extreme values of ff on the boundary of the domain
  3. Compare all of these ff values - the largest is the absolute max, the smallest is the absolute min
Example
Find the absolute max and min values of the function f(x,y)=x2y+y2f(x,y)=x^2y+y^2 on the closed triangular region with vertices (1, 0), (0, 1), and  (−1, 0)(1,\ 0),\ \left(0,\ 1\right),\ \text{and }\ \left(-1,\ 0\right)
1. We first need to find the critical points
fx=2xy=0f_x=2xy=0
fy=x2+2y=0f_y=x^2+2y=0
The only point that satisfies both of these equations is (0,0)\left(0,0\right), so this is the only critical point.
f(0,0)=0f\left(0,0\right)=0

2. Now we need to find the max and min on the boundary of the domain, which consists of the 3 lines in the following graph:
On line 1:
  • The max value of ff is when x=0, y=1x=0,\ y=1 → f(0,1)=1f\left(0,1\right)=1
  • The min value of ff is when x=1, y=0x=1,\ y=0 → f(1,0)=0f\left(1,0\right)=0
On line 2:
  • The y=0y=0 and −1≤x≤1-1\le x\le1, so the max and min value of ff on this line is 0
One line 3:
  • The max value of ff is when x=0x=0, y=1y=1 → f(0,1)=1f\left(0,1\right)=1
  • The min value of ff is when x=−1, y=0x=-1,\ y=0 → f(−1,0)=0f\left(-1,0\right)=0

3. Comparing these values, we see that the absolute max is 1 at the point (0,1)\left(0,1\right) and the absolute min is 0 at (1,0), (−1,0)  and  (0,0)\left(1,0\right),\ \left(-1,0\right)\ \text{ and }\ \left(0,0\right).
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Lagrange Multipliers

To find the max and min values of the function f(x,y)f\left(x,y\right) subject to the constraint g(x,y)=0g\left(x,y\right)=0:
  1. Write a new function F(x,y,λ)=f(x,y)+λ g(x,y)F\left(x,y,\lambda\right)=f\left(x,y\right)+\lambda\ g\left(x,y\right)
  2. Find ∇F=⟨Fx, Fy, Fz⟩\nabla F=\left\langle F_x,\ F_y,\ F_z\right\rangle
  3. Solve for points that satisfy Fx=0,  Fy=0,  Fλ=0F_x=0,\ \ F_y=0,\ \ F_{\lambda}=0 → these are the critical points
  4. Classify these critical points as max or min
Example
Find the maximum value of f(x,y)=x+yf\left(x,y\right)=x+y subject to the constraint x2+y2=32x^2+y^2=32
First we need to rewrite the constraint as x2+y2−32=0x^2+y^2-32=0, now we can use the method of Lagrange Multipliers:
1. F(x,y,λ)=x+y+λ(x2+y2−32)F\left(x,y,\lambda\right)=x+y+\lambda\left(x^2+y^2-32\right)

2. ∇F=⟨1+2λx,  1+2λy,  x2+y2−32⟩\nabla F=\left\langle1+2\lambda x,\ \ 1+2\lambda y,\ \ x^2+y^2-32\right\rangle

3. We need to solve {1+2λx=01+2λy=0x2+y2−32=0\begin{cases} 1+2\lambda x=0\\ 1+2\lambda y=0\\ x^2+y^2-32=0 \end{cases}
From the first equation: x=−12λx=-\frac{1}{2\lambda}
From the second equation: y=−12λy=-\frac{1}{2\lambda}
Substitute both of these into the third equation:
14λ2+14λ2−32=0\frac{1}{4\lambda^2}+\frac{1}{4\lambda^2}-32=0
12λ2=32\frac{1}{2\lambda^2}=32
λ2=164\lambda^2=\frac{1}{64}
λ=±18\lambda=\pm\frac{1}{8}
So, the corresponding critical points are (4,4)\left(4,4\right) and (−4,−4)\left(-4,-4\right)

4. f(4,4)=8f\left(4,4\right)=8 and f(−4,−4)=−8f\left(-4,-4\right)=-8

Therefore, the max value is 8 and it occurs at (4,4)\left(4,4\right), and the min value is −8-8 and it occurs at (−4, −4)\left(-4,\ -4\right)

Practice: Lagrange Multipliers

You want to maximize the volume of an opened-top rectangular based cardboard box. You only have 100 cm2 of cardboard available. Which of the following is a mathematical model that represents this problem?

Practice: Lagrange Multipliers

Which of the following is true about f(x,y)=2x−yf\left(x,y\right)=2x-y subject to the constraint x2+y2=20x^2+y^2=20?