I had no idea what the "Nelder and Mead method" was until I looked it up on the internet.
What I would do to "Find the stationary points of the function \(\displaystyle f(x,y)= x^2+ 3(y- 1)^4\)" is to set the "gradient" equal to 0. The gradient is \(\displaystyle grad f= \nabla f= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}= 2x\vec{i}+ 12(y- 1)^3\vec{j}\). Setting that equal to 0 gives 2x= 0 and \(\displaystyle 12(y- 1)^3= 0\) so that x= 0 and y= 1. The only stationary point is (0, 1). The "Hessian" mentioned is the determinant \(\displaystyle \left|\begin{array}{cc}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial x\partial y} & \frac{\partial^2 f}{\partial y^2} \end{array} \right|=\)\(\displaystyle \left(\frac{\partial^2 f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right)_{edited}-\left(\frac{\partial^2 y}{\partial x\partial y}\right)^2= 2(12)- 0= 24\). Since that is positive the stationary point is a minimum.
Without being that technical, we could observe that \(\displaystyle x^2\) is a parabola opening upward while \(\displaystyle 3(y- 1)^4\) also opens upward. Again that tells us that the stationary point is a minimum.
