Mathematics Nonlinear Optimization Solution 2022

Running head: MATHEMATICS 1
Nonlinear Optimization
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Student Number
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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)