Optimization Problems Solution: Mathematics Module 2, AY 2019-2020

This document contains the solutions to three exercises related to optimization problems. Exercise 1 explores finding stationary points, Hessian matrices, and determining whether a point is a minimum, maximum, or neither. Exercise 2 focuses on optimization with constraints, involving the Lagrangian function and identifying minimum points. Exercise 3 delves into optimization problems with inequality constraints, utilizing the Weierstrass theorem, Kuhn-Tucker conditions, and finding global maximums. The solutions provide detailed steps, explanations, and analysis of the mathematical concepts involved, including the application of theorems and conditions to solve the given optimization scenarios.

SOLUTIONS
EXERCISE ONE
(a)
In this case g: R² R, we shall only look for stationary points
∆g (x, y) =0 dg/dx(x,y)=4x+2y=0
dg/dy(x,y)=2y+2x=0
We can divide by 2, moreover the difference in (i) and (ii) yields x=0
Our system has two equations
(i) 2x+y=0
X=0 (0,0) (-1,2)
(ii) X + y=0
X=0 ( 0, 0) (-1,1)
(b)
Hessian matrix Hg(x,y)
2 1
1 1
G is a convex function has two different systems.
(c)
(0, 0) saddle point
(-1, 2) Local Maximum
(-1, 1) Local minimum

EXERCISE TWO
(a)
The Lagrangian function is then given by
L (α, x) =2x²+2xy+y²+ α (1-x-y)
Setting ∆L=0 means
L (1-x-y) =0 (i)
4x+2y- α =0 (ii)
2x+2y- α =0 (iii)
The difference (ii) – (iii) yields
(4x-2x) + (2y-2y) +(- α -- α) =0
That is 2x=0, x=0
(i) 1-x-y=0
1-y=0
Given that x=0 and y=1 then; P (0, 1,1)
(b)
Insert α =1
L (1, x) =2x²+2xy+y²+1-x-y
4x+2y-1
2x+2y-1

2x+y=1
x+ y=1
x=2
y=-1
(2, -1) Local minimum
(c)
No point of global maximum because we have indefinite form.
EXERCISE THREE
(a)
2x²+2xy+y²≤4
By completing square method, we get
Centre as (-1, -1)
A is complete, the objective function f is continuous, hence waitress theorem
(Essed, H.,2014) guarantees the existence of the global maximum.

Since g∑R² and g= dg/dx(4x+2y)dg/dy(2y+2x) centre (-1, -1) where the
constant is not binding, Karush-Kuhn-Tucker condition holds and the
constraints is everywhere regular.
(b)
The Lagrangian function is
L (α, x) = x + y+ α (4-2x²-2xy-y²),
(c)
L¹x=0 1-4 α x-2 α y=0 (i)
L¹y=0 1-2 α x-2 α y=0 (ii)
L¹ α ≥0 α (4-2x²-2xy-y²) =0 (iii)
4-2x²-2xy-y²≥0 (iv)
α ≥0 α ≥0 (v)
From (v) we get either α =0 or α >0
If α =0
x=y=0
(d)
Constraints not binding hence
If α >0X²=¾X=√¾
We have two points (0,0) (√¾, √¾)
Since f (0, 0) =0
F (√¾, √¾) =¾ Global maximum.
References

Essed, H., & Therese, W. (2014). The harassed waitress problem. In
International Conference on Fun with Algorithms (pp. 325-339).