Let's solve the problem **step by step** as instructed: --- ### **Step 1: Understanding the Problem** - **Function:** \( f(x, y) = x^2 + (y - 1)^2 \) - **Constraint:** \( y = |x| \) - **Task:** Mark the minimum and maximum of \( f(x, y) \) **on the given figure**. - **No algebraic computation!** - **Sketch level curves** of \( f \) to help. - **Explain if min/max does not exist.** --- ### **Step 2: Sketch Level Curves (Contours) of \( f(x, y) \)** - The function \( f(x, y) = x^2 + (y-1)^2 \) is a family of circles centered at \( (0, 1) \) in the \( xy \)-plane. - **Level curve** for value \( c \): \( x^2 + (y-1)^2 = c \) This is a circle centered at \( (0, 1) \). - **Sketch:** Draw a few circles centered at \( (0, 1) \) with increasing radii. --- ### **Step 3: Draw the Constraint Curve \( y = |x| \)** - This is a "V" shape, opening upward, meeting at the origin. --- ### **Step 4: Where Does the Constraint Intersect the Level Curves?** - The possible points are **only where \( y = |x| \)**. - **The closest point on \( y = |x| \) to (0, 1)** will be the minimum. - The **furthest points, as \( x \to \pm\infty \), \( y \to \infty \)**, will give the maximum (if it exists). --- ### **Step 5: Find the Minimum (Graphically)** - The **minimum** occurs where the circle is smallest and still touches the "V" curve. - The **point on the "V" closest to (0, 1)** is **(0, 1)**. - This is because: - At \( x = 0 \), \( y = |0| = 0 \). So (0, 0) is on the constraint. - But at \( x = 0 \), \( y = 1 \) is the center of the circles, but (0, 1) is **not** on the constraint. - Let's check, **what's the closest point on \( y = |x| \) to (0, 1)?** - Draw a perpendicular from (0,1) to the "V". The closest point is where the vertical line through \( x = 0 \) hits the "V", which is at (0, 0). - But let's check if moving along the "V" gets closer: For \( x > 0 \), the point is \( (x, x) \), the distance to (0, 1) is \( \sqrt{x^2 + (x-1)^2} \). - For \( x = 0.5 \), the point is (0.5, 0.5), distance to (0, 1) is \( \sqrt{0.25 + 0.25} = \sqrt{0.5} \approx 0.707 \). - For \( x = 1 \), (1, 1), distance = 1. - For \( x = 0 \), (0, 0), distance = 1. - The minimum occurs at \( x = 0.5 \), \( y = 0.5 \), since that is closer to (0, 1). - More generally, the closest point happens at the intersection of the constraint and the line joining (0, 1) to the constraint. - **But graphically**, the closest point is somewhere **above the vertex**, along both arms, symmetrically. - **Conclusion:** The minimum occurs at two points, one on each arm of the "V", equidistant from (0, 1). --- ### **Step 6: Find the Maximum** - As \( x \to \infty \), the points on the "V" move away from (0, 1) indefinitely. - The circles get larger and larger. - **There is NO maximum**; the function increases without bound. --- ### **Step 7: Final Answers on the Figure** 1. **Sketch a few circles centered at (0, 1).** 2. **Mark the minimum as the points on the "V" that are closest to (0, 1)**. These are the two points on each arm of the "V" at the same distance from (0, 1). 3. **State that no maximum exists**, as the function increases without bound along the arms of the "V". --- ## **Final Answer (Summary):** - **Minimum**: At the points on \( y = |x| \) **closest to (0, 1)** (one on each arm), which are approximately at \( (0.5, 0.5) \) and \( (-0.5, 0.5) \). - **Maximum**: **Does not exist**; the function increases without bound as you move out along either arm of the "V". - **Level curves**: Draw several circles centered at (0, 1). - **Mark the minimum points and state "No maximum" on the figure.** --- **(You can now sketch this on your paper according to the instructions!)** --- **If you need a sketch, let me know!**

Certainly! Here's a detailed, step-by-step solution with explanations, including the reasoning behind each part, without performing algebraic computations, as per the instructions. --- ### **Problem Restatement:** - **Function:** \( f(z, y) = z^2 + (y - 1)^2 \) - **Constraint:** \( y = |z| \) - **Task:** Find the locations of the **minimum and maximum** of \( f(z, y) \) **subject to** the constraint \( y = |z| \). - **Note:** Do **not** perform algebraic calculations; instead, use geometric intuition and sketch level curves to determine extrema. --- ### **Step 1: Understand the Function and Constraint** - \( f(z, y) = z^2 + (y-1)^2 \) is a sum of squares: - **Level curves** of \( f \): circles centered at \( (0, 1) \) in the \( zy \)-plane. - **Center:** at point \( (0, 1) \). - The **constraint \( y = |z| \)**: - Graphically, this is a "V" shape opening upward, with vertex at the origin \( (0, 0) \). - For positive \( z \), \( y = z \); for negative \( z \), \( y = -z \). --- ### **Step 2: Sketch Level Curves of \( f(z, y) \)** - The level curves are circles centered at \( (0, 1) \). - As the value of \( c \) (level of \( f \)) increases, these circles grow larger. - The smallest circle (approaching zero) is at the point \( (0, 1) \). - Visualize several circles with increasing radii, centered at \( (0, 1) \). --- ### **Step 3: Sketch the Constraint \( y = |z| \)** - Draw the "V" shape: - The vertex at \( (0, 0) \). - The arms: \( y = z \) (for \( z \geq 0 \)), and \( y = -z \) (for \( z \leq 0 \)). --- ### **Step 4: Find the Min and Max of \( f(z, y) \) on the Constraint** - **Goal:** Identify points on \( y = |z| \) where \( f(z, y) \) is minimized and maximized. --- ### **Step 5: Geometric Intuition for the Minimum** - Since \( f(z, y) = z^2 + (y-1)^2 \), the value depends on: - Distance from \( (z, y) \) to \( (0, 1) \). - To **minimize** \( f \), find the point on the "V" closest to \( (0, 1) \): - The **closest point** on the "V" to \( (0, 1) \) is the point with **smallest distance**. - The **point on the "V"** that is closest to \( (0, 1) \): - Is **somewhere along the arms** of the "V" near the vertex at \( (0, 0) \). - Since \( (0, 1) \) is directly above the vertex, the **closest points** on the arms are roughly at \( (z, y) \) with \( y = |z| \), **near** the vertex. - **Expected location of the minimum**: - On the arms at points **closest to** \( (0, 1) \), which geometrically are at a certain finite distance from \( (0, 1) \). --- ### **Step 6: Intuition for the Maximum** - As \( |z| \to \infty \), \( y = |z| \to \infty \), so: - The points move **far away** along the arms of the "V". - The value of \( f(z, y) \) becomes **very large** because \( z^2 \) and \( (y-1)^2 \) both grow large. - **Therefore**, \( f(z, y) \to \infty \) as \( |z| \to \infty \). - **Conclusion**: - There is **no maximum value** of \( f \) on the constraint; it is **unbounded above**. --- ### **Step 7: Final Markings on the Sketch** - **Minimum point**: - Located **on the arms of the "V"**, roughly where the level curves (circles centered at \( (0, 1) \)) are tangent or closest to the "V". - These points are **symmetrically** located on each arm at some finite distance from the vertex. - **Maximum**: - **Does not exist** because \( f \) increases without bound as you move along the arms away from the vertex. --- ### **Summary of the Results:** | Aspect | Description | |----------------------------|---------------------------------------------------------------| | **Minimum** | At points on \( y = |z| \) closest to \( (0, 1) \). Approximately at some finite \( z \) near the vertex, symmetric on both arms. Mark these points on your sketch. | | **Maximum** | **Does not exist**; \( f(z, y) \to \infty \) as \( |z| \to \infty \). No maximum point exists. | --- ### **Additional Tips for Sketching:** - Draw concentric circles centered at \( (0, 1) \) to represent level curves. - Draw the "V" shape \( y = |z| \). - Mark the points on the arms of the "V" that are closest to \( (0, 1) \) — these are the minimum points. - Indicate that as you move further along the arms, the function increases indefinitely, so no maximum. --- **This completes the detailed, step-by-step reasoning without algebraic calculations.**