z=f(x,y)=x^2+2y^2-2x-8y+5在条件x^2+2y^2

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/15 05:25:02

先把x,y换成参数坐标然后目标函数就变成z=(Cos[a] - 1)^2 + 2 (Sin[a]/Sqrt[2]- 2)^2 - 4然后就可以求出极值：为了便于理解附上图一张

Plot[(Cos[a] - 1)^2 + 2 (Sin[a]/Sqrt[2] - 2)^2 - 4, {a, -2 Pi, 2 Pi}]
接下就可以求极值
FindMinimum[(Cos[a] - 1)^2 + 2(Sin[a]/Sqrt[2] - 2)^2 - 4, {a, 0, -2 Pi, 2 Pi}] 求最小值出来为{0., {a -> 1.23096}} FindMaximum[(Cos[a] - 1)^2 + 2(Sin[a]/Sqrt[2] - 2)^2- 4, {a, 0, -2 Pi, 2 Pi}]求出来最大值为 {12., {a -> -1.91063}}括号中的a为当a取该值时取到该极值

分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y) (4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai 1.求函数f(x,y,z)=x^2+y^2+z^2在限制条件x+y+z=1下的最小值试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) x/2=y/3=z/5 x+3y-z/x-3y+z (y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z x,y,z为实数且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-2y)^2+(x+y x,y,z为实数且(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y- 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(x-2z+y)(y+z-2x)+(x (x+y-z)^2-(x-y+z)^2=? 设Z=X+Y,其中X,Y满足X+2Y>=0,X-Y x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)