電子科大實(shí)驗(yàn)報(bào)告之四:罐裝飲料瓶(Matlab的解法)
題目:
設(shè)易拉罐的側(cè)面的厚度1單位,
s=2*PI*r*h+4*PI*r^2
h高度,r為半徑,體積V=PI*h*r^2.
V一定時(shí),求使s最小的r和h.
Matlab的解法:
解法1:
v=350;
s=0;h=0;r=0;
for j=1:6
?r_=(v/(j*pi))^(1/3)
?h_=j*r_
?s_=2*pi*r_*h_+4*pi*r_*r_
?
?if(s>s_ | s==0)
??s=s_;
??h=h_;
??r=r_;
?end
end
s
r
h
r/h
解法2:
v=350;
s=[];h=[];r=[];
for j=1:6
?r(j)=(v/(j*pi))^(1/3)
?h(j)=j*r(j)
?s(j)=2*pi*r(j)*h(j)+4*pi*r(j)^2
end
[minCost,j]=min(s)
r(j)
h(j)
r(j)/h(j)
解法3:
cost=inline('(2*pi*x+4*pi)*(350/(pi*x))^(2/3)');
s=fminbnd(cost,0.001,25)
feval(cost,s)
fplot(cost,[0.1 50]);grid on;