(x²-2y²+2x+2x+2)e^(x-y) B=f"xy=(-x²+2y²-2x-4y)e^(x-y) C=f"yy=(x²-2y²+4y+4y-4)e^(x-y) 在(0, 0), A=2, B=0,C=-4, B²-AC=8>0, 不是极值点; 在(-4, -2), A=(16-8-8-8+2)e^(-2)=-6e^(-2) B=(-16+8+8+8)e^(-2)=8e^...
百度试题 结果1 题目二次型 f(x,y)=2x2-xy-y2的系数矩阵是___。相关知识点: 试题来源: 解析 答案: 证明题反馈 收藏
=[(ⅹ+y)/2][(x+y)/2][1-(ⅹ+y)]≤[(2(x+y)/2+1-(ⅹ+y))/3]³=1/27.∴(ⅹ+y)/2=1-(x+y)且x=y，即x=y=1/3时，所求f(x,y)|max=1/27。
由z=f（xy，2x+y 2 ），得z x =yf′ 1 +2f′ 2 ，z y =xf′ 1 +2yf′ 2∴由全微分公式dz=z x dx+z y dy，得dz=（yf′ 1 +2f′ 2 ）dx+（xf′ 1 +2yf′ 2 ）dy
x= 1 2(u+v), y= 1 2(u−v),于是由f(x+y,x-y)=4(x2-xy-y2),得f(u,v)=4uv-u2+v2,故f(x,y)=4xy-x2+y2,xf'x(x,y)+yf'y(x,y)=x(4y-2x)+y(4x+2y)=-2x2+8xy+2y2,故选:D. 令x+y=u,x-y=v,利用多元函数偏导数的概念即可求出. 本题考点:多元函数偏导数的概念...
={[(xy)'(x^2-y^2)+(xy)(x^2-y^2)'](x^2+y^2)-xy(x^2-y^2)(2xdx+2ydy)}/(x^2+y^2)^2={[(ydx+xdy)(x^2-y^2)+xy(2xdx-2ydy)](x^2+y^2)-2xy(x^2-y^2)(xdx+ydy)}/(x^2+y^2)^2=【y(x^4-4x^2y^2-y^4)/(x^2+y^2)^2】dx-【x(x^4-5y^4)/(x...
百度试题 结果1 题目已知函数f(x+y,xy)=x2+y2,则0f(x,y),f(x,y)oxay=( ) A. 2x+2y; B. 2x-2; C. 2x-2y; D. 2x+2. 相关知识点: 试题来源: 解析 B 反馈 收藏
A = fxx = -2 < 0, B = fxy = 1, C = fyy = -2,AC - B^2 = 3 > 0, (-2, -2) 是极大值点， 极大值 f(-2, -2) = 8 另解 f(x,y)=xy-x^2-y^2-2x-2y+4 = -(x-y/2+1)^2 - (3/4)(y+2)^2 + 8, 最大值 f(-2, -2) = 8 ...
f'x=y(2-x-y)-xy=y(2-2x-y)=0,得y=0或y=2-2xf'y=x(2-x-y)-xy=x(2-x-2y)=0,得x=0或x=2-2y解得有以下几组(0,0), (2,0), (0, 2), (2/3,2/3)A=f"xx=-2yB=f"xy=2-2x-2yC=f"yy=-2x在(0,0), AC-B^2=0-2^2=-4 ...
解答解:x2+y2-xy+2x-y+1=0, ∴4x2+4y2-4xy+8x-4y+4=0, 化为3(x2+2x+1)+(x2+4y2+2x-4xy-4y+1)=0, 即3(x+1)2+(x-2y+1)2=0, ∴{x+1=0x−2y+1=0{x+1=0x−2y+1=0, 解得{x=−1y=0{x=−1y=0. ∴x=-1,y=0. ...