---
title: "Extreme value of x²+y²+z² with x+y+z=1"
subject: "Mathematics"
chapter: 1
key: "semester_2/mathematics:extreme-value-x2-y2-z2-3"
source: https://www.bctengineer.com/q/semester_2/mathematics/extreme-value-x2-y2-z2-3
site: BCT Engineering
---

# Extreme value of x²+y²+z² with x+y+z=1

Mathematics · Chapter 1: Calculus of Two or more variables

- Asked in: 2067 Chaitra, 5 marks
- Video: [Maximum and Minimum of f(x,y,z) using  Lagrange's Multipliers](https://www.youtube.com/watch?v=xTkLsaO509M) — TEAM Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/extreme-value-x2-y2-z2-3
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-1

## Question

Find the extreme value of x²+y²+z² subject to the condition x+y+z = 1.

## Short answer

Introduce Lagrange multiplier λ: 2x+λ=0, 2y+λ=0, 2z+λ=0 implies x=y=z. Constraint x+y+z=1 gives x=y=z=1/3. The function is squared distance from origin to plane; minimum value is 1/3 at (1/3,1/3,1/3).

## Answer

Find the extreme value of f = x² + y² + z² subject to x + y + z = 1.

#### Lagrange conditions
F = x² + y² + z² + λ(x + y + z − 1)

```text
2x + λ = 0
2y + λ = 0
2z + λ = 0
```

So x = y = z = −λ/2. The constraint gives 3x = 1, so x = y = z = 1/3 (λ = −2/3).

#### Value
f = 3 × (1/3)² = 3 × 1/9 = 1/3.

#### Nature
f is the squared distance from the origin to a point of the plane x + y + z = 1. This distance has a minimum (foot of the perpendicular, the point (1/3, 1/3, 1/3), at distance 1/√3) but no maximum. Also, by Cauchy–Schwarz, 1 = (x + y + z)² ≤ 3(x² + y² + z²), so f ≥ 1/3.

**Answer: minimum value = 1/3, at (1/3, 1/3, 1/3).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Extremising x²+y²+z² with x+y+z = 1 gives:
   A. (1,0,0)
   B. (1,1,1)
   C. (1/3, 1/3, 1/3)
   D. (1/2,1/2,0)
   Answer: C. 3x = 1. (Key point: Stationary point)

2. Value there:
   A. 3
   B. 1
   C. 1/3
   D. 1/9
   Answer: C. 3×(1/9) = 1/3. (Key point: Value)

3. The value is a minimum since:
   A. λ negative
   B. f is linear
   C. It is squared distance to a plane, with Cauchy–Schwarz giving f ≥ 1/3
   D. the plane is bounded
   Answer: C. 1 = (x+y+z)² ≤ 3f. (Key point: Nature)

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