Using optfun command to solve lagrange multipliers

Question

0 个投票

hello,

I was given this code to run an optfun command, and I am a little confused how the optfun function works.

function F = optfun(pair)

x = pair(1);

y = pair(2);

F = [3*y+.001*y*exp(x)-4*x, x^2+y^2-25];

end

Any help would be great! I need to solve a lagrange multiplier with this.

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Answer 1

Koushik Kureti 2020-3-12

0 个投票

Hello,

‘optfun’ is function name taking input arguments ‘pair’. Return value of the function is stored in F.

X is assigned with first value of pair. Y is assigned with second value of pair.

Now F has two values, where first value is 3*y+.001*y*exp(x)-4*x and second value are x^2+y^2-25.

At the end F is returned. You can call the function by ‘optfun’ (pair) where defining the pair before calling.

Example:

pair = [1 2];
disp(optfun(pair));
 
function F = optfun(pair)
x = pair(1);
y = pair(2);
F = [3*y+.001*y*exp(x)-4*x, x^2+y^2-25];
end

Output:

2.0054 -20.0000