用matlab 求一元函数最值y=1/5*r*acos((r^2-1/2*((5325683144035133/2305843009213693952*r^2+21/5050*(50500*r^2-r^4)^(1/2)-9/5650*(56500*r^2-r^4)^(1/2))^2+(-1478907458814005/576460752303423488*r^2-4/2525*(50500*r^
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[email protected](r) 1/5*r*acos((r^2-1/2*((5325683144035133/2305843009213693952*r^2+21/5050*(50500*r^2-r^4)^(1/2)-9/5650*(56500*r^2-r^4)^(1/2))^2+(-1478907458814005/576460752303423488*r^2-4/2525*(50500*r^2-r^4)^(1/2)+11/2825*(56500*r^2-r^4)^(1/2))^2)^(1/2))*r^2)*(1+exp(10-1/10*r^2))+1/5*((80-4/2525*r^2+21/5050*(50500*r^2-r^4)^(1/2))^2+(210-21/5050*r^2-4/2525*(50500*r^2-r^4)^(1/2))^2)^(1/2)+1/5*((-220+11/2825*r^2-9/5650*(56500*r^2-r^4)^(1/2))^2+(-90+9/5650*r^2+11/2825*(56500*r^2-r^4)^(1/2))^2)^(1/2);
[a,b]=fminbnd(y,-1,1)
算出来的a就是在[-1 1]区间上的最小值点
b是对应的最小值