Using optfun command to solve lagrange multipliers

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.

回答 (1 件)

Koushik Kureti
Koushik Kureti 2020 年 3 月 12 日

0 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

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2019 年 10 月 15 日

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