Question

x^2/4+(y+1)^2/9+(z+1)^2/16=1,则2x+y+z-16最大值和最小值

Answer 1

#include<stdio.h>

void main()

{

 double x,y,z,min=0,max=0,sum;

 int i=0;

 

 for(x=-2;x<=2;x+=1)

 for(y=-4;y<=2;y+=1)

 for(z=-5;z<=3;z+=1)

    if((x*x/4+(y+1)*(y+1)/9+(z+1)*(z+1)/16)==1)

 {

   sum=2*+y+z-16;

   printf("x=%g\ty=%g\tz=%g\t",x,y,z);

   if(i==0){max=sum;min=sum;i++;}

   if(sum>max)max=sum;

   if(min>sum)min=sum;

   printf("i=%d\tmax=%g\tmin=%g\n\n\n",i,max,min);

 }

 printf("max=%g\tmin=%g\n",max,min);

}

Answer 2

解：由题设及柯西不等式可知，41=(4²+3²+4²)·{(x/2)²+[(y+1)/3]²+[(z+1)/4]²}≥(2x+y+1+z+1)².===>|2x+y+z+2|≤√41.等号仅当x²=64/41,(y+1)²=81/41,(z+1)²=256/41时取得。故(2x+y+z-16)max=(√41)-18, (2x+y+z-16)min=(-√41)-18.