The solution is L1 = 3.33 m, L2 = 6.67 m with Lambda = -0.5547.
The value of lambda is negative in this problem.
The F90 code is shown below
program main
real :: u,L1,L2, lambda, denom1, denom2
integer :: maxiter
real,allocatable :: a(:,:)
real,allocatable :: b(:)
! Problem statement
! Assume the top pole locations are at (0,0,10) and (x2,y2,5)
! where x2,y2 are user's defined constant
! The constraint are
! x2**2+y2**2+25 = 100 (distance between poles are 10 m)
! Objective function
u(L1,L2) = SQRT(25.0+L1**2)+SQRT(100.+L2**2)+lambda*(L1+L2-10.0)
allocate(a(3,3))
allocate (b(3))
maxiter = 5
L1 = 5.0 ; L2 = 5.0 ; lambda = 1.
do i = 1,maxiter
denom1 = SQRT(25.0+L1**2)
denom2 = SQRT(100.0+L2**2)
a(1,1) = (denom1-L1**2/denom1)/denom1**2; a(1,2) = 0. ;a(1,3) = 1.
a(2,1) =0.; a(2,2) = (denom2-L2**2/denom2)/denom2**2; a(2,3) = 1.0
a(3,1) = 1.0; a(3,2) = 1.0; a(3,3) = 0.0
b(1) = -(L1/denom1 + lambda)
b(2) = -(L2/denom2+lambda)
b(3) = -(L1+L2-10.)
! solve for delta_L1, delta_L2, delta_lambda
! from GaussElimination subroutine, the solution is store in b vector
CALL GaussElimination(a, b, 3)
L1 = L1 + b(1)
L2 = L2 + b(2)
lambda = lambda + b(3)
print*,'L1=',L1,'L2=',L2,'lambda=',lambda,'U=',u(L1,L2)
end do
end program
