Calculating max, min, and saddle points

As of right now, I'm dealing with an assignment which wants to calculate the max, min, and saddle point of a given function. On the bottom is my current code. The problem that I am facing is I am unable to create a loop where my eqA_double are able to read X and Y solutions from M.

if true
  % code
end
function Local_ex(x,y,z)
syms x
syms y
a = ((y.*x^3)+(x.*y^3)-(1.69.*x.*y));
eqA_partial_diff_x = diff(a,x)
eqA_partial_diff_y = diff(a,y)
M = solve(eqA_partial_diff_x,eqA_partial_diff_y)
N = [M.x,M.y]
eqA_double_partial_diff_x = diff(eqA_partial_diff_x,x)
eqA_double_partial_diff_y = diff(eqA_partial_diff_y,y)
eqA_double_partial_diff_y_then_x = diff(eqA_partial_diff_y,x)

Answer 1

[madhan ravi]

    syms x y

    a = ((y.*x^3)+(x.*y^3)-(1.69.*x.*y));
    eqA_partial_diff_x = diff(a,x)
    eqA_partial_diff_y = diff(a,y)
    M = solve(eqA_partial_diff_x,eqA_partial_diff_y)
    N = [M.x,M.y]
    [m,n]=size(N)

    eqA_double_partial_diff_x = diff(eqA_partial_diff_x,x)
    eqA_double_partial_diff_y = diff(eqA_partial_diff_y,y)
    eqA_double_partial_diff_y_then_x = diff(eqA_partial_diff_y,x)

        x=N(:,1)
        y=N(:,2)

            eqA_double_partial_diff_x=subs(eqA_double_partial_diff_x)

    eqA_double_partial_diff_y =subs(eqA_double_partial_diff_y)

    eqA_double_partial_diff_y_then_x=subs(eqA_double_partial_diff_y_then_x)

%Determinant part

syms x y

a = ((y.*x^3)+(x.*y^3)-(1.69.*x.*y));
eqA_fx = diff(a,x)
eqA_fy = diff(a,y)
M = solve(eqA_fx,eqA_fy)
N = [M.x,M.y]
eqA_fxx = diff(eqA_fx,x)
eqA_fyy = diff(eqA_fy,y)
eqA_fyx = diff(eqA_fy,x)

D = eqA_fxx.* eqA_fyy - eqA_fyx^2

subs(D,{x,y},{N(:,1),N(:,2)})

Remark: You don’t need loop to read the values one by one instead you can substitute all the values at once

[Muhammad Waseem]

I realized and just used sub, but I'm confused on how to sub in the values for the discriminant

if true
  % code
end
function Local_ex(x,y,z)
syms x
syms y
a = ((y.*x^3)+(x.*y^3)-(1.69.*x.*y));
eqA_fx = diff(a,x)
eqA_fy = diff(a,y)
M = solve(eqA_fx,eqA_fy)
N = [M.x,M.y]
eqA_fxx = diff(eqA_fx,x)
eqA_fyy = diff(eqA_fy,y)
eqA_fyx = diff(eqA_fy,x)
subs(eqA_fxx,{x,y},{M.x(1),M.y(1)})
subs(eqA_fxx,{x,y},{M.x(2),M.y(2)})
subs(eqA_fxx,{x,y},{M.x(3),M.y(3)})
subs(eqA_fxx,{x,y},{M.x(4),M.y(4)})
subs(eqA_fxx,{x,y},{M.x(5),M.y(5)})
subs(eqA_fxx,{x,y},{M.x(6),M.y(6)})
subs(eqA_fyy,{x,y},{M.x(1),M.y(1)})
subs(eqA_fyx,{x,y},{M.x(1),M.y(1)})
subs(eqA_fyx,{x,y},{M.x(2),M.y(2)})
subs(eqA_fyx,{x,y},{M.x(3),M.y(3)})
subs(eqA_fyx,{x,y},{M.x(4),M.y(4)})
subs(eqA_fyx,{x,y},{M.x(4),M.y(4)})
D = eqA_fxx.* eqA_fyy - eqA_fyx^2
sub(D,{x,y},{M.x(1),M.y(1)})

[Muhammad Waseem]

i think i got it lol

if true
  % code
end
function Local_ex(x,y,z)
syms x 
syms y
a = ((y.*x^3)+(x.*y^3)-(1.69.*x.*y));
eqA_fx = diff(a,x)
eqA_fy = diff(a,y)
M = solve(eqA_fx,eqA_fy)
N = [M.x,M.y]
[m,n]=size(N)
eqA_fxx = diff(eqA_fx,x)
eqA_fyy = diff(eqA_fy,y)
eqA_fyx = diff(eqA_fy,x)
d = eqA_fxx.*eqA_fyy - eqA_fyx^2
    x=N(:,1)
    y=N(:,2)
eqA_fxx = subs(eqA_fxx)
eqA_fyy = subs(eqA_fyy)
eqA_fyx = subs(eqA_fyx)
Answer_D= subs(d)

[madhan ravi]

Yes you got the point.See my edited answer above.

[Muhammad Waseem]

Thank you so much. Honestly thank you.