---
title: "Golden section search for a minimum"
subject: "Machine Design II"
chapter: 2
key: "semester_7/machine_design_ii:golden-section-search-minimum"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/golden-section-search-minimum
site: BCT Engineering
---

# Golden section search for a minimum

Machine Design II · Chapter 2: Optimization Techniques

- Asked in: year not given, 3+5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/golden-section-search-minimum
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-2

## Question

(a) Explain the golden section search method for a single-variable function, and state why it is better than the exhaustive search. (b) Use the golden section method to locate the minimum of $f(x) = x^2 + 54/x$ in the interval $[1, 5]$. Carry out four iterations and give the final interval and the estimated minimum. Take the golden ratio constant as 0.618.

## Short answer

Golden section search places two interior points at 0.618 intervals, reuses one point, needing only one new evaluation per reduction—far more efficient than exhaustive search. For f(x)=x²+54/x on [1,5], after four iterations the interval is [2.89,3.47]; estimated minimum near x≈3.2 (true x=3).

## Answer

#### (a) Golden section search
It is an interval-reduction method for a **unimodal** function (one minimum in the interval). Two interior points are placed at

$$
x_1 = b - 0.618(b-a), \qquad x_2 = a + 0.618(b-a)
$$

- If $f(x_1) < f(x_2)$, the minimum lies in $[a, x_2]$, so set $b = x_2$.
- Otherwise it lies in $[x_1, b]$, so set $a = x_1$.

The old interior point is reused as one new interior point, so each iteration needs **only one new function evaluation** and shrinks the interval by 0.618. Exhaustive search needs many evaluations for the same accuracy, so golden section is much cheaper. It needs no derivative.

#### (b) Numerical
Initial $[a,b] = [1,5]$, length 4.

| Iter | $a$ | $b$ | $x_1$ | $x_2$ | $f(x_1)$ | $f(x_2)$ | Keep |
|---|---|---|---|---|---|---|---|
| 1 | 1.000 | 5.000 | 2.528 | 3.472 | 27.752 | 27.608 | $[x_1, b]$ |
| 2 | 2.528 | 5.000 | 3.472 | 4.056 | 27.608 | 29.763 | $[a, x_2]$ |
| 3 | 2.528 | 4.056 | 3.111 | 3.472 | 27.036 | 27.608 | $[a, x_2]$ |
| 4 | 2.528 | 3.472 | 2.889 | 3.111 | 27.038 | 27.036 | $[x_1, b]$ |

After four iterations the interval is $[2.889, 3.472]$ (length 0.583, which is $4\times0.618^4$).

Estimated minimum at the mid-point: $x^* \approx 3.18$, $f \approx 27.09$.

**Check by calculus:** $f' = 2x - 54/x^2 = 0 \Rightarrow x^3 = 27 \Rightarrow x = 3$, $f_{min} = 27$. The search interval contains this point.

**Answer:** final interval $[2.89,\ 3.47]$; $x^* \approx 3.2$ (true minimum $x=3$, $f=27$).

## Check yourself

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

1. What property must the function have in the interval for golden section search?
   A. It must be linear
   B. It must have several equal minima
   C. It must be unimodal, with a single minimum in the interval
   D. It must have a known derivative
   Answer: C. Interval reduction works only if there is one minimum in the interval, and no derivative is needed. (Key point: Unimodal requirement)

2. In golden section search, if f(x1) < f(x2), which interval is kept?
   A. [a, x2], so set b = x2
   B. [x1, b], so set a = x1
   C. [x1, x2]
   D. The whole interval [a, b]
   Answer: A. If f(x1) is smaller the minimum lies on the left, in [a, x2]. Otherwise it lies in [x1, b]. (Key point: Interval reduction rule)

3. Why is golden section cheaper than exhaustive search?
   A. It uses derivatives
   B. The old interior point is reused, so each iteration needs only one new function evaluation and the interval shrinks by 0.618
   C. It evaluates the function at every point in the interval
   D. It guarantees the exact minimum in one step
   Answer: B. Reusing one interior point means one new evaluation per iteration while the interval shrinks by 0.618 each time. (Key point: Advantage over exhaustive search)

4. For f(x) = x^2 + 54/x on [1, 5], x1 = 2.528 gives 27.752 and x2 = 3.472 gives 27.608. What is the new interval?
   A. [1, 3.472]
   B. [1, 2.528]
   C. [2.528, 3.472]
   D. [2.528, 5]
   Answer: D. f(x1) is greater than f(x2), so the minimum lies in [x1, b] = [2.528, 5]. The true minimum is x = 3. (Key point: First iteration)

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