方程p=1/(0.6-a)+1/(0.66-b)+1/(0.72-c)
0<a<0.6;
0<b<0.66;
0<c<0.72;
能用matlab求一个P的最大值吗？能绘图？
你好，同学，这是一个非线性规划问题
p=@(x)1/(0.6-x(1))+1/(0.66-x(2))+1/(0.72-x(3));
x = fmincon(@(x)-p(x),[2;2;2],[],[],[],[],[0;0;0],[0.6;0.66;0.72]);
a = x(1)
b = x(2)
c = x(3)
pmax = p(x)
a =
0.599999999830908
b =
0.656231453913813
c =
0.712199711985334
pmax =
5.913926508040196e+09