---
title: "Extreme values with two constraints"
subject: "Mathematics"
chapter: 1
key: "semester_2/mathematics:extreme-values-two-constraints"
source: https://www.bctengineer.com/q/semester_2/mathematics/extreme-values-two-constraints
site: BCT Engineering
---

# Extreme values with two constraints

Mathematics · Chapter 1: Calculus of Two or more variables

- Asked in: 2071 Magh, 5 marks; 2068 Magh, 5 marks (asked 2 times)
- Video: [Two constraints - Lagrange Multipliers - Find the extreme values of the function: 2x-3y+z, g=x-2x+2z](https://www.youtube.com/watch?v=z3H1dNbwPac) — Calculus Physics Chem Accounting Tam Mai Thanh Cao
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/extreme-values-two-constraints
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-1

## Question

Find the extreme values of x²+y²+z² subjected to the condition x+y+z−1 = 0 and xyz+1 = 0.

## Short answer

Using Lagrange multipliers for x²+y²+z² with x+y+z=1 and xyz=−1 gives stationary points (1,1,−1) and its permutations. The value is 3, which is the minimum; there is no maximum.

## Answer

Find the extreme values of f = x² + y² + z² subject to g₁ = x + y + z − 1 = 0 and g₂ = xyz + 1 = 0.

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

```text
2x + λ + μyz = 0
2y + λ + μxz = 0
2z + λ + μxy = 0
```

Subtract the first two: 2(x − y) − μz(x − y) = 0, so (x − y)(2 − μz) = 0.
Similarly (y − z)(2 − μx) = 0 and (z − x)(2 − μy) = 0.

#### Case analysis
- If x, y, z were all different, then μz = μx = μy = 2, forcing x = y = z, a contradiction. So at least two are equal.
- Take x = y. Then z = 1 − 2x and x²z = −1:

```text
x²(1 − 2x) = −1
2x³ − x² − 1 = 0
(x − 1)(2x² + x + 1) = 0
```

2x² + x + 1 has discriminant 1 − 8 < 0 (no real root), so x = 1, giving y = 1, z = −1.

The stationary points are (1, 1, −1), (1, −1, 1), (−1, 1, 1), and at each

x² + y² + z² = 1 + 1 + 1 = 3.

#### Nature
Since x + y + z = 1, x²+y²+z² = 1 − 2(xy + yz + zx). The numbers x, y, z are roots of t³ − t² + e t + 1 = 0 where e = xy + yz + zx. All three roots are real only if its discriminant −(e + 1)(4e² − 5e + 23) ≥ 0, i.e. e ≤ −1. Hence x² + y² + z² = 1 − 2e ≥ 3, so 3 is the minimum, and there is no maximum (the value can be made as large as we like).

**Answer: the extreme value is the minimum 3, attained at (1, 1, −1) and its permutations.**

## Check yourself

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

1. With two constraints, how many multipliers does the Lagrange function use?
   A. One, λ
   B. Three
   C. Two, λ and μ
   D. None
   Answer: C. One multiplier per constraint: F = f + λ g₁ + μ g₂. (Key point: Setup)

2. Subtracting the first two equations gives (x−y)(2−μz) = 0. What conclusion follows if x, y, z were all different?
   A. μz = μx = μy = 2 forcing x = y = z, a contradiction
   B. μ = 0
   C. x + y + z = 0
   D. z = 0
   Answer: A. All three factors would vanish, giving x = y = z, contradicting that they are different. So at least two are equal. (Key point: Case analysis)

3. Taking x = y, z = 1−2x and x²z = −1, what is the real solution?
   A. x = 1, giving (1, 1, −1)
   B. x = −1, giving (−1, −1, 3)
   C. x = 1/2
   D. x = 2
   Answer: A. 2x³ − x² − 1 = (x−1)(2x²+x+1); the quadratic has no real root, so x = 1, z = -1. (Key point: Solving for x)

4. What is the final result for x²+y²+z²?
   A. Minimum 1
   B. Maximum 3
   C. Minimum 3, no maximum
   D. Both maximum and minimum equal 3
   Answer: C. Real roots force e ≤ −1, so 1−2e ≥ 3; the value can be arbitrarily large. (Key point: Nature)

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