Question

33–40 find the absolute maximum and minimum values of f on the set d. 33. ( f(x, y) = x^2 + y^2 - 2x ), d is the closed triangular region with vertices (2, 0), (0, 2), and (0, −2)

Explanation:

Step1: Find critical points in the interior of D

First, find the partial derivatives of \( f(x,y)=x^{2}+y^{2}-2x \).
The partial derivative with respect to \( x \) is \( f_{x}=2x - 2 \), and with respect to \( y \) is \( f_{y}=2y \).
Set \( f_{x}=0 \) and \( f_{y}=0 \):
\( 2x - 2 = 0\Rightarrow x = 1 \)
\( 2y = 0\Rightarrow y = 0 \)
So the critical point is \( (1,0) \), and \( f(1,0)=1^{2}+0^{2}-2\times1=- 1 \).

Step2: Analyze the boundaries of D

Boundary 1: Line segment from \( (2,0) \) to \( (0,2) \)

The equation of this line is \( x + y=2 \), \( x\in[0,2] \), \( y = 2 - x \).
Substitute \( y = 2 - x \) into \( f(x,y) \):
\( f(x,2 - x)=x^{2}+(2 - x)^{2}-2x=x^{2}+4 - 4x+x^{2}-2x = 2x^{2}-6x + 4 \)
Take the derivative with respect to \( x \): \( f^\prime(x)=4x-6 \)
Set \( f^\prime(x)=0 \), we get \( x=\frac{3}{2} \), then \( y=2-\frac{3}{2}=\frac{1}{2} \)
\( f(\frac{3}{2},\frac{1}{2})=2\times(\frac{3}{2})^{2}-6\times\frac{3}{2}+4=2\times\frac{9}{4}-9 + 4=\frac{9}{2}-9 + 4=-\frac{1}{2} \)
Evaluate at endpoints:
\( f(2,0)=2^{2}+0^{2}-2\times2 = 0 \)
\( f(0,2)=0^{2}+2^{2}-2\times0 = 4 \)

Boundary 2: Line segment from \( (2,0) \) to \( (0,-2) \)

The equation of this line is \( x - y=2 \), \( x\in[0,2] \), \( y=x - 2 \)
Substitute \( y=x - 2 \) into \( f(x,y) \):
\( f(x,x - 2)=x^{2}+(x - 2)^{2}-2x=x^{2}+x^{2}-4x + 4-2x=2x^{2}-6x + 4 \)
Take the derivative with respect to \( x \): \( f^\prime(x)=4x - 6 \)
Set \( f^\prime(x)=0 \), we get \( x = \frac{3}{2} \), then \( y=\frac{3}{2}-2=-\frac{1}{2} \)
\( f(\frac{3}{2},-\frac{1}{2})=2\times(\frac{3}{2})^{2}-6\times\frac{3}{2}+4=-\frac{1}{2} \)
Evaluate at endpoints:
\( f(2,0)=0 \)
\( f(0,-2)=0^{2}+(-2)^{2}-2\times0 = 4 \)

Boundary 3: Line segment from \( (0,2) \) to \( (0,-2) \)

Here \( x = 0 \), \( y\in[-2,2] \)
Substitute \( x = 0 \) into \( f(x,y) \):
\( f(0,y)=0^{2}+y^{2}-2\times0=y^{2} \)
The minimum value of \( y^{2} \) on \( y\in[-2,2] \) is \( 0 \) (when \( y = 0 \)) and the maximum value is \( 4 \) (when \( y=\pm2 \))

Step3: Compare all function values

We have the following function values:

• Critical point: \( f(1,0)=-1 \)
• Boundary 1: \( f(2,0)=0 \), \( f(0,2)=4 \), \( f(\frac{3}{2},\frac{1}{2})=-\frac{1}{2} \)
• Boundary 2: \( f(2,0)=0 \), \( f(0,-2)=4 \), \( f(\frac{3}{2},-\frac{1}{2})=-\frac{1}{2} \)
• Boundary 3: \( f(0,0)=0 \), \( f(0,\pm2)=4 \)

Answer:

The absolute minimum value of \( f \) on \( D \) is \(-1\) (at the point \((1,0)\)) and the absolute maximum value is \(4\) (at the points \((0,2)\) and \((0, - 2)\)).