---
title: "Maximum of xyz with x+y+z=24"
subject: "Mathematics"
chapter: 1
key: "semester_2/mathematics:maximum-xyz-x-y-z"
source: https://www.bctengineer.com/q/semester_2/mathematics/maximum-xyz-x-y-z
site: BCT Engineering
---

# Maximum of xyz with x+y+z=24

Mathematics · Chapter 1: Calculus of Two or more variables

- Asked in: 2075 Baisakh, 5 marks; 2073 Magh, 5 marks; 2071 Bhadra, 5 marks; 2069 Bhadra, 5 marks (asked 4 times)
- Video: [Lagrange Multipliers Maximum of f(x, y, z) = xyz subject to x + y + z - 3 = 0](https://www.youtube.com/watch?v=JHvpTE1Jo_U) — The Math Sorcerer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/maximum-xyz-x-y-z
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-1

## Question

Obtain the maximum value of xyz such that x+y+z = 24.

## Short answer

Using Lagrange multipliers for xyz with x+y+z=24, stationarity gives x=y=z=8. The maximum value is 512.

## Answer

Maximise f = xyz subject to g = x + y + z − 24 = 0 using Lagrange's multipliers (x, y, z > 0).

#### Setting up
Let F = xyz + λ(x + y + z − 24). Stationary conditions:

```text
∂F/∂x = yz + λ = 0
∂F/∂y = xz + λ = 0
∂F/∂z = xy + λ = 0
x + y + z = 24
```

#### Solving
From the first three equations, yz = xz = xy = −λ.

- yz = xz gives z(y − x) = 0, so x = y (z ≠ 0).
- xz = xy gives x(z − y) = 0, so y = z.

Hence x = y = z. Using the constraint, 3x = 24, so x = y = z = 8 (and λ = −64).

#### Nature of the point
The constraint set with x, y, z > 0 is a bounded triangle, and xyz → 0 on its boundary while xyz > 0 inside, so the single stationary point must be a maximum. (Check with AM–GM: (xyz)^(1/3) ≤ (x+y+z)/3 = 8, so xyz ≤ 512.)

**Answer: maximum value of xyz = 8 × 8 × 8 = 512, at x = y = z = 8.**

## Check yourself

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

1. Which auxiliary function is used to maximise xyz subject to x+y+z = 24?
   A. F = xyz + λ(x + y + z)
   B. F = xyz − 24 + λ
   C. F = xyz + λ(x + y + z − 24)
   D. F = x + y + z + λ(xyz − 24)
   Answer: C. The constraint g = x+y+z-24 = 0 is multiplied by λ and added to the objective xyz. (Key point: Lagrange function)

2. From yz + λ = 0, xz + λ = 0, xy + λ = 0 (x, y, z > 0), what follows?
   A. x + y = z
   B. x = 0
   C. x = y = z
   D. y = z = 0
   Answer: C. yz = xz = xy = -λ; cancelling the nonzero factors gives x = y and y = z. (Key point: Solving stationary equations)

3. Using the constraint, what are the values of x, y, z at the stationary point?
   A. x = y = z = 6
   B. x = y = z = 8
   C. x = y = z = 12
   D. x = y = z = 24
   Answer: B. 3x = 24 gives x = 8, so x = y = z = 8. (Key point: Stationary point)

4. Why is the single stationary point a maximum?
   A. Because λ is negative
   B. Because the second derivative of xyz is zero
   C. xyz → 0 on the boundary of the triangle while xyz > 0 inside
   D. Because x, y, z are all equal
   Answer: C. On the bounded triangle xyz vanishes at the boundary and is positive inside, so the only interior stationary point must be the maximum. The maximum value is 512. (Key point: Why a maximum)

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