Question

identify any extrema of the function by recognizing its given form or its form after completing the square. verify your results by using the partial derivatives to locate any critical points and test for relative extrema. (if an answer does not exist, enter dne.)
g(x,y)=(x - 2)^2+(y - 3)^2
relative minimum (x,y,z)=()
relative maximum (x,y,z)=()

Explanation:

Step1: Analyze the function form

The function \(g(x,y)=(x - 2)^2+(y - 3)^2\) is in the form of the sum of two squares. Since \((x - 2)^2\geq0\) and \((y - 3)^2\geq0\) for all real \(x\) and \(y\), the minimum value of \(g(x,y)\) occurs when \(x - 2=0\) and \(y - 3=0\).
When \(x = 2\) and \(y=3\), \(g(2,3)=(2 - 2)^2+(3 - 3)^2=0\).
For any other values of \(x\) and \(y\), \((x - 2)^2>0\) or \((y - 3)^2>0\), so \(g(x,y)>0\).
Since the function is a sum of non - negative terms, it has no maximum (as \(x\to\pm\infty\) or \(y\to\pm\infty\), \(g(x,y)\to+\infty\)).

Step2: Use partial derivatives (verification)

First, find the first - order partial derivatives:
The partial derivative with respect to \(x\) is \(g_x(x,y)=2(x - 2)\)
Set \(g_x(x,y)=0\), then \(2(x - 2)=0\), which gives \(x = 2\)
The partial derivative with respect to \(y\) is \(g_y(x,y)=2(y - 3)\)
Set \(g_y(x,y)=0\), then \(2(y - 3)=0\), which gives \(y = 3\)
The second - order partial derivatives: \(g_{xx}(x,y)=2\), \(g_{yy}(x,y)=2\), \(g_{xy}(x,y)=0\)
The discriminant \(D=g_{xx}(x,y)g_{yy}(x,y)-[g_{xy}(x,y)]^2=(2)(2)-0^2 = 4>0\) and \(g_{xx}(2,3)=2>0\)

Answer:

relative minimum \((x,y,z)=(2,3,0)\); relative maximum: DNE