Spring 2020 Math 410: Homework on Nonlinear Optimization Techniques

This document presents complete solutions to a Math 410 homework assignment focusing on nonlinear optimization techniques. The assignment covers several key areas, including the application of Lagrange multipliers to solve constrained optimization problems, such as minimizing and maximizing functions subject to equality constraints. It also explores the Kuhn-Tucker conditions, used to solve optimization problems with inequality constraints, determining the optimal values of variables and the associated Lagrange multipliers. Furthermore, the assignment delves into the method of feasible directions, providing step-by-step iterations to solve a maximization problem. The solutions provide detailed explanations, including the determination of critical points, the satisfaction of necessary and sufficient conditions, and an analysis of how changes in constraints affect the optimal function values. The document is a valuable resource for students studying optimization methods in calculus and related fields.

Running head: MATHEMATICS 1
Nonlinear Optimization
Name
Student Number
Institution

MATHEMATICS 2
Solutions
Question 1
Minimize: f(x, y) =3x2 + y2 + 2xy + 6 x + 2 y
s.t. 2x-y=4.
F(x, y, λ) = f(x, y)= f(x, y)- λg((x, y)
=3x2 + y2 + 2xy + 6 x + 2 y-λ (2x-y-4)
Introducing Lagrange’s multiplier;
F x=6 x +2 y +6−2 λ−−−−−−−−−−−−−−−−−−−−−−−−−−(1)
F y =2 y +2 x+ 2 — λ−−−−−−−−−−−−−−−−−−−−−−−−−−(2)
Fλ=2 x− y =4−−−−−−−−−−−−−−−−−−−−−−−−−−−−(3)
From equation 1, we can get the value of λ i.e.
6x+2y+6 = 2λ
λ = 3x+y+3
In equation 2, λ = 2y+2x+2
Thus, equating λ we get;
3x+y+3 = 2y+2x+2
x+y+1 = 0

MATHEMATICS 3
y = x+1
Substituting in the equation 3, we get;
2x-x-1=4
x=5
y=6
λ = 3x+y+3 = 24
Question 2
Maximize f(x, y, w) =x y w
s.t. 2x + 3y + 4w =36
Let f(x, y = x y w
g((x, y) = 2x + 3y + 4w -36
By defining a new function with Lagrange’s multiplier, we get;
F(x, y, λ) = f(x, y)= f(x, y)- λg((x, y)
F(x, y, λ) = x y w- λ(2x + 3y + 4w -36)
To find the critical numbers of F, we set the partial derivatives of F with respect to x, y, w, and λ
i.e.
F x= yw−2 λ−−−−−−−−−−−−−−−−−−−−−−−−−−(1)
F y =xw — 3 λ−−−−−−−−−−−−−−−−−−−−−−−−−−(2)
F w=xy −4 λ−−−−−−−−−−−−−−−−−−−−−−−−−−(3)

MATHEMATICS 4
Fλ=2 x +3 y+ 4 w−36−−−−−−−−−−−−−−−−−−−−( 4)
We therefore solve for x in the first equation and substitute in the second equation;
yw=2 λ
λ= yw
2
Putting it in equation 2 i.e. xw-3( yw
2 ¿ =0
y = 2
3 x
Putting the solution in equation 3, we get;
xy−4( yw
2 ¿=0
w = x
2
Putting the functions in equation 4, we get the values x i.e.
2 x+3 ¿ 2
3 x)+ 4 ( x
2 )=36
6 x=36
x=6
y = 2
3 x = 4
w = x
2 = 3

MATHEMATICS 5
¿ yw
2 =6
Question 3
Minimize z= (x-1)2 + (y-2)2
s.t -x + y < 1
x+ y < 2
x, y > 0
By defining a new function with Lagrange’s multiplier, we get;
F(x, y, λ) = f(x, y)= f(x, y)- λg((x, y)
We can therefore find the Kuhn-Tucker conditions as follows;
L = ( x−1 ) 2 + ( y−2 )2 −λ 1 ( −x + y −1 ) + λ 2 ( x + y−2 )
Lx ¿ 2 ¿1) + λ 1+ λ 2=0−−−−−−−−−−−−−−−−−−−−−−(1)
Ly = 2 ¿1) - λ 1+λ 2=0−−−−−−−−−−−−−−−−−−−−−−−−(2)
L λ 1=−x+ y −1−−−−−−−−−−−−−−−−−−−−−−−−−−−(3)
Lλ 1=x + y −2−−−−−−−−−−−−−−−−−−−−−−−−−−−−− ( 4 )
The above four equations attains the following conditions;
Lx ¿ 2 ¿1) + λ 1+ λ 2=0 x > 0
Ly = 2 ¿1) - λ 1+λ 2=0 y > 0

MATHEMATICS 6
Lλ 1=−x+ y ≤ 1
Lλ 2=x+ y ≤ 2
The four equations contain four unknown variables i.e. x , y , λ 1 ,∧λ 2
To find these unknowns, we substitute equation 1 to equation 4 in the following procedures;
2 ( x−1 )=−λ 1−λ 2
x=− ( λ 1+ λ 2
2 )+1
y= ( λ 1−λ 2
2 )+ 2
Equating the above solutions to equation 4, we get;
λ 2=0∧ λ1=4, y=4 , and x=−3
Question 4
Maximize z= x y - x2 – y2 + 5 x + 6y
s.t 3x + y < 4
x, y > 0
By defining a new function with Lagrange’s multiplier, we get;
F(x, y, λ) = f(x, y)= f(x, y)- λg((x, y)
We can therefore find the Kuhn-Tucker conditions as follows;

MATHEMATICS 7
L=xy −x2 − y2+ 5 x +6 y−¿ λ (3 x + y −4)
L x= y−2 x+5−3 λ−−−−−−−−−−−−−−−−−−−−−−(1)
L y=x−2 y +6+λ−−−−−−−−−−−−−−−−−−−−−−−( 2)
Lλ=3 x+ y ≤ 4−−−−−−−−−−−−−−−−−−−−−−−−−−(3)
Making λ the subject in equation 1, we get;
λ= 1
3 ( y−2 x +5 )
Again, making λ in equation 2 the subject, we get;
λ=x−2 y +6
We then equate the two equations for λ ¿ get ;
1
3 ( y−2 x +5 )=x −2 y +6
Making x the subject, we get;
x=5 y +23
Substituting the value of x in equation 3, we get;
3 ( 5 y+ 23 ) + y=4
y=−65
16 , x = 43
16 ,∧λ=−71
48
Question 5

MATHEMATICS 8
Maximize f(x, y) = 2 x y – x2 – y2 + 5 x + 6 y
st. 2 x + y < 4
x, y > 0
Start at (0,0)
We first find the gradient which will be at;
∂ f ( x , y )
∂ x =2 y−2 x+5
∂ f ( x , y )
∂ y =2 x−2 y+ 6
Gradient is at (5, 6) since the region is marked to stat at ( 0,0 )
If we use the condition constraint i.e. 2 x + y < 4 and we already know the value of x and y, we
get that 2(5) +6 < 4 is not true thus violates the constraint for the feasible direction.
We will then have the following iteration;

MATHEMATICS 9
From the drawing above, feasible directions includes those vectors that are within the gradients
while non feasible directions are those that violate the constraint function i.e. where 2 x + y < 4
and the values are greater than 4.