1. [1/2 Points] DETAILS MY NOTES LARCALC9 13.8.001. Identily any extrema of the function by recognizing its given form or its form after completing the square. extrema. $$ g(x, y)=(x-2)^{2}+(y-5)^{2} $$

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We have the function

$$
g(x, y) = (x-2)^2 + (y-5)^2.
$$

**Step 1. Recognize the Form**

The function is in the form of the sum of two squared differences, which represents the squared distance from the point $$(2,5)$$ in the plane.

**Step 2. Identify the Minimum**

Since any square is non-negative, we have

$$
(x-2)^2 \geq 0 \quad \text{and} \quad (y-5)^2 \geq 0.
$$

Thus,

$$
g(x, y) \geq 0.
$$

The function attains its minimum value when both squared terms equal zero. This happens when

$$
x - 2 = 0 \quad \text{and} \quad y - 5 = 0,
$$

which gives

$$
x=2 \quad \text{and} \quad y=5.
$$

At the point $$(2,5)$$, the value of the function is

$$
g(2,5)= (2-2)^2 + (5-5)^2 = 0+0 = 0.
$$

**Step 3. Discuss the Maximum**

Since $$(x-2)^2$$ and $$(y-5)^2$$ can become arbitrarily large as $$x$$ or $$y$$ moves away from $$2$$ or $$5$$, the function has no upper bound. Therefore, $$g(x, y)$$ does not have a maximum value.

**Final Answer**

- The function has a global minimum of $$0$$ at $$(x,y)=(2,5)$$.
- There is no maximum value for the function.
The function $$ g(x, y) = (x-2)^2 + (y-5)^2 $$ has a minimum value of 0 at the point (2,5). There is no maximum value for this function.

1. [1/2 Points] DETAILS MY NOTES LARCALC9 13.8.001. Identily any extrema of the function by recognizing its given form or its form after completing the square. extrema. \[ g(x, y)=(x-2)^{2}+(y-5)^{2} \]

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