Question

this is an optimization subject.

that is example 2.33

Question 2 (6 Marks) (Chapter 2) Consider the function f : R3 -R defined as f(x1,2,3 +4eli++21), (G) Explain why f has a global minimum over the set Hint: Read Example 2.33 (i) Find the global minimum point and global minimum value of f over the set C. Example 2.33. Consider the function/(x1,x2)=xf+xỈ over the set The set C is not bounded, and thus the Weierstrass theorem does not guarantee the exis- tence of a global minimizer of f over C, but since f is coercive and C is closed, Theorem 2.32 does guarantee the existence of such a global minimizer. It is also not a difficult task to find the global minimum point in this example. There are two options: In one op- tion the global minimum point is in the interior of C, and in that case by Theorem 2.6 Vf(x) = 0, meaning that x = o, which is impossible since the zeros vector is not in C. The other option is that the global minimum point is attained at the boundary of C given by bd(C) (x,x2): xx2--1). We can then substitute x-x2-1 into the objective function and recast the problem as the one-dimensional optimization problem of mini- mizing g (%) = (-1-x2)2 + xỈ over R. Since g,(ね) = 2(1 + x21+ 2x2, it follows that g, has a single root, which is =-05, and hence x,--0.5. Since (x1,x2)-(-05,-05) is the only candidate for a global minimum point, and since there must be at least one global minimizer, it follows that (x1,x2) = (-0.5,-0.5) is the global minimum point of/ over C. I Question 2 (6 Marks) (Chapter 2) Consider the function f : R3-R defined as f(1,2, 3)3 +4eli+3+21-1) (G) Explain why f has a global minimum over the set C: {x E R3 x,-x,-1). : Hint: Read Example 2.33 () Find the global minimum point and global minimum value of f over the set C.

Answer 1

Since we are finding minimum over the set $C=\{x\in\mathbb R^3:x_1-x_3=1\}$ , we can substitute $x_1-1$ for $x_3$ in the expression of the objective function, to get

$\begin{align*}f(x)&=3x_1^2+4x_2^4+e^{x_1^2+x_3^2+2x_1-1}\\ &=3x_1^2+4x_2^4+e^{x_1^2+x_1^2-2x_1+1+2x_1-1}\\ &=3x_1^2+4x_2^4+e^{2x_1^2}\end{align*}$

Since each summand above is non-negative, we conclude that $\begin{align*}f(x)&=3x_1^2+4x_2^4+e^{2x_1^2}\end{align*}$ is minimum when the summands $\begin{align*}3x_1^2,4x_2^4,e^{2x_1^2}\end{align*}$ are simultaneously minimum. Note that the minimum of each of these is attained at $\begin{align*}x_1=0=x_2\end{align*}$ . This corresponds to
r = (2. 1.2.2, 2.1-1) = (0.01-1)
, which is a point of global minimum of the function
f (r)
. This proves that global minimum exists.

The value at the global minimum is

- 0+0+ e

The point of global minimum is .

(0.0.-1)