Find function's minimum for specific constant values

Question

Valeria 2018-6-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/403975-find-function-s-minimum-for-specific-constant-values

评论： Walter Roberson 2018-6-5

I have a function mew which is a derivative of another function, fun:

syms v a b
fun = -a*v + (2*b*v^(2/3))/(exp(0.5) - 2*(v^(1/3))*log(exp(0.5)*v^(1/3)));
mew=diff(fun,v)

I must find mew's minimum for a=0, b=1.3

How?

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Answer 1

Walter Roberson 2018-6-4

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https://ww2.mathworks.cn/matlabcentral/answers/403975-find-function-s-minimum-for-specific-constant-values#answer_323169

在 MATLAB Online 中打开

G = subs(mew, [a, b], [0, 1.3]);
sol = solve(G);
mewmin = subs(G, v, sol)
crosscheck = subs(G, v, sol+1)

The reason for doing the crosscheck is that solve() is only able to find a single numeric root, and without further checking you do not know whether it will be a maxima or a minima. The fact that the crosscheck comes out with a larger value (less negative) than the min gives evidence that we are indeed working with a minima not a maxima. But you should really look at sign( subs(diff(mew), v, sol ) to be sure.

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Walter Roberson 2018-6-4

Note that if there are multiple minima then this will not find them all to check which is the global minimum.

Walter Roberson 2018-6-5

在 MATLAB Online 中打开

MATLAB finds sol as

6396480385065521911547770910773962672562070583/(1427247692705959881058285969449495136382746624*lambertw(0, (3445831591435597800619981388529*exp(-1))/1267650600228229401496703205376)^3)

However, if you change the exp(0.5) to exp(sym(0.5)) then the more accurate solution is

exp(3/2)/LambertW(1)^3

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Answer 2

Sid Parida 2018-6-4

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https://ww2.mathworks.cn/matlabcentral/answers/403975-find-function-s-minimum-for-specific-constant-values#answer_323173

在 MATLAB Online 中打开

Try this:

You can play wit the lb, and ub to get the global minimum:

syms v a b
fun = -a*v + (2*b*v^(2/3))/(exp(0.5) - 2*(v^(1/3))*log(exp(0.5)*v^(1/3)));
mew = diff(fun,v);
subbed_ab = matlabFunction(subs(mew, {a, b}, {0.0, 1.3}));
lower_bound = -100000;
upper_bound =  100000;
min_val= fminbnd(subbed_ab, lower_bound, upper_bound);