Statics - Pendulum and Spring System

I have attached simulation code below. The code sweeps the angle and looks for the angle which minimizes the net torque on the rod. I have also attached a plot of torque vs angle.

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import math

###########################

# Constants

dtheta = (math.pi/2.0)/(10.0**6.0)

L = 0.75
L0 = L/5.0
g = 9.81
m = 2.0
k = 20.0

Bx = 2.0*L
By = 0.0

###########################

theta = dtheta

Tmin = 999999999.0

while theta <= math.pi/2.0 - dtheta:

    rx = L*math.sin(theta)   # coordinates of rod end
    ry = -L*math.cos(theta)

    rgx = rx/2.0   # lever arm coordinates for gravity
    rgy = ry/2.0

    Fgx = 0.0     # gravity force
    Fgy = -m*g

    Dx = Bx - rx  # displacement vector from rod end to B
    Dy = By - ry

    D = math.hypot(Dx,Dy)  

    ux = Dx/D     # unitized version of D vector
    uy = Dy/D

    s = D - L0    # spring stretch

    Fs = k*s      

    Fsx = Fs * ux  # spring force 
    Fsy = Fs * uy

    Tg = rgx*Fgy - rgy*Fgx  # gravity torque
    Ts = rx*Fsy - ry*Fsx    # spring torque

    T = Tg + Ts  # net torque

    if math.fabs(T) < Tmin:            # store theta value for zero net torque
        Tmin = math.fabs(T)
        theta_store = theta

    #print (theta*180.0/math.pi),T

    theta = theta + dtheta

###########################

print ""
print ""

print Tmin
print (theta_store * 180.0 / math.pi)

# Results
#1.08662274618e-05
#68.3400599997

The length of the stretched spring, $l$ , is obtained from the equation $l^2=(2L-L\sin \theta) ^2+L^2\cos^2 \theta$ , or $l=L\sqrt {5-4\sin \theta}$ . The force exerted on the lower end of the rod along the spring is, therefore, given by $F=kL(\sqrt {5-4\sin \theta}-\dfrac{1}{5})$ . The angle between the rod and the spring, $α$ , is given by $\sin α=\dfrac{2\cos \theta}{\sqrt {5-4\sin \theta}}$ . Balancing moments of the weight of the rod and the spring force about the top end of the rod we get $196046.68976656\sin^6 \theta-487258.8972164\sin^5 \theta-51450.61123975\sin^4 \theta+879485.3728\sin^3 \theta-385924.216\sin^2 \theta-396800\sin \theta+246016=0$ . The two positive real solutions of this equation are $\theta=68.3401\degree$ and $\theta=74.598\degree$ . The value satisfying the conditions of the problem is $\boxed {68.3401\degree}$