We will using the so called slack variables to transform inequality into equality relations. So considering s as a slack variable, the initial problem
Find local extrema for
f(x,y) = e^-x+e^(-3y)-x y subject to x+2y<7
is transformed into an equivalent one
f(x,y) = e^-x+e^(-3y)-x y subject to g(x,y,s) = x+2y+ s^2 -7 = 0
Forming the lagrangian
L(x,y,s,lambda) = f(x,y) + lambda(x,y,s)
The local extrema are included in the lagrangian stationary points. This is true mainly because L(x,y,s,lambda) is analytical.
The stationary points are determined finding the solutions to
grad F(x,y,s,lambda) = vec 0
The resulting equations are
{
(-e^-x + lambda - y=0), (-3 e^(-3 y) + 2 lambda - x=0), (2 lambda s=0), (-7 + s^2 + x + 2 y=0)
:}
This system of equations must be solved numerically using an iterative procedure like Newton-Raphson.
https://en.wikipedia.org/wiki/Newton%27s_method
Calling X = {x,y,s,lambda} and F(X) the vector of equations the iterative procedure is
X_{k+1} = X_k - kappa H(X_k)^{-1} F(X_k)
where H(X) = grad(grad F(X))
H(X) =
((e^-x, -1, 0, 1),(-1, 9 e^(-3 y), 0, 2),(0, 0, 2 lambda, 2 s),(1, 2,
2 s, 0))
kappa is a convergence factor here choosed as 1
After some tries we find
X_0 = {3.52143,1.73929,-3.33992*10^-20,1.76884} as a stationary point. This point is at the restriction boundary because s = -3.33992*10^-20 approx 0
The qualification is made over
f(x,g(x,y,0)=0) =(f_g)(x) = e^(3/2 (x-7)) + e^-x + 1/2 ( x-7) x
Calculating (d^2f_g)/(dx^2) for x = 3.52143 we have
(d^2f_g)/(dx^2) = 1.04175 > 0 characterizing the stationary point as a local minimum.
Attached a figure with the contour map and the soution point
![enter image source here]()
