Line minimisation techniques Question 3Let f: R2R be defined by f(x)= 3x - 2x₁x₂ + x² + 4x₂ + 3x₂.(a) Find be R² and Q R2x2 such that Q> 0 and f(x) = x¹Qx+b¹x.(b) Find the stationary point & of f(x). Show that x is a local minimiser of f(x).(c) Minimise f(x) using 1 iteration of the method of Hooke and Jeeves with starting pointXo = [0,0], stepsize h₁ = 1 in the e₁ = [1,07 direction and stepsize h₂ = 1 in thee₂ = [0, 1] direction.

Answered: Let : R2R be defined by f(x)= 3x-2x₁₂ +…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Line minimisation techniques

Question 3
Let f: R2R be defined by f(x)= 3x - 2x₁x₂ + x² + 4x₂ + 3x₂.
(a) Find be R² and Q R2x2 such that Q> 0 and f(x) = x¹Qx+b¹x.
(b) Find the stationary point & of f(x). Show that x is a local minimiser of f(x).
(c) Minimise f(x) using 1 iteration of the method of Hooke and Jeeves with starting point
Xo = [0,0], stepsize h₁ = 1 in the e₁ = [1,07 direction and stepsize h₂ = 1 in the
e₂ = [0, 1] direction.

Expert Solution

Step 1: Solving part (a) through the introduction of Quadratic forms

Let $f : R^{2} \to R$ be defined by $f (x) = 3 x_{1}^{2} - 2 x_{1} x_{2} + x_{2}^{2} + 4 x_{1} + 3 x_{2}$ .

(a)

The quadratic term $Q \in R^{2 \times 2}$ can be obtained from the terms $3 x_{1}^{2} - 2 x_{1} x_{2} + x_{2}^{2}$ .

We know that,

$1 half x to the power of T Q x equals 1 half open square brackets table row cell x subscript 1 end cell cell x subscript 2 end cell end table close square brackets open square brackets table row a b row c d end table close square brackets open square brackets table row cell x subscript 1 end cell row cell x subscript 2 end cell end table close square brackets equals 1 half open square brackets table row cell x subscript 1 end cell cell x subscript 2 end cell end table close square brackets open square brackets table row cell a x subscript 1 plus b x subscript 2 end cell row cell c x subscript 1 plus d x subscript 2 end cell end table close square brackets equals 1 half open parentheses a x subscript 1 superscript 2 plus b x subscript 1 x subscript 2 plus c x subscript 1 x subscript 2 plus d x subscript 2 superscript 2 close parentheses equals a over 2 x subscript 1 superscript 2 plus fraction numerator open parentheses b plus c close parentheses over denominator 2 end fraction x subscript 1 x subscript 2 plus d over 2 x subscript 2 superscript 2$

Comparing with $3 x_{1}^{2} - 2 x_{1} x_{2} + x_{2}^{2}$ , we get

$a over 2 equals 3 comma space fraction numerator b plus c over denominator 2 end fraction equals negative 2 comma space d over 2 equals 1 a equals 6 comma space b plus c equals negative 4 comma space d equals 2.$