trying to find all critical points where gradient f = 0 but am getting an error.

This is what my code looks like right now. I am getting an error with the 'Real' part of my code with it saying that it is invalid.

%Storing the function f and g

syms w x y z;
f(w,x,y,z) = 2.0*w*x - 4.0*y - 5.0*w*y + 4.0*y*z + w^2 + 4.0*z^2;
g(w,x,y,z) = 0.4723*w*x + 0.6455*w*y + 0.1187*x*y + 0.8042*w^2 + 0.08831*x^2 + 0.2568*y^2 + 0.06482*z^2;

%Getting the derivatives of both functions
Dfw = diff(f,w);
Dfx = diff(f,x);
Dfy = diff(f,y);
Dfz = diff(f,z);

Dgw = diff(g,w);
Dgx = diff(g,x);
Dgy = diff(g,y);
Dgz = diff(g,z);

%Finding All Critical points by finding where gradient f is zero

[wSols, xSols, ySols, zSols] = solve([Dfw == 0, Dfx == 0, Dfy == 0, Dfz == 0], [w,x,y,z],'Real', 'true')

Answer 1

[Walter Roberson]

Use 'Real', true instead of 'Real', 'true'

[Peter Pyrka]

Okay it seems to goes through. I end up getting this as the output:

wSols =

0

xSols =

-5

ySols =

-2

zSols =

1

The lab i am trying to complete says that this command i do should print wSols,xSols, etc. as column vectors containing all possible solutions to the system. Is this output that I got correct and agree with that?

[Walter Roberson]

Both of those systems have partial derivatives that are strictly linear. There is only a single solution for each of them.

By the way: have a look at gradient()

[Peter Pyrka]

Okay but just to clarify, my output should be correct then with having only 1 solution for each and not more since it is strictly linear?

[Walter Roberson]

There would have to be a subexpression with degree 3 or higher for the partial derivatives to be non-linear, so having only one solution is expected for these.