己知x-y≤1,2x y≤5,x≥|,没z=3x y,求z的最大值及相应点坐标
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/29 12:27:15
先把可行域画出来x^2+y^2就是离原点的距离的平方,所以最大值在离原点最远的地方取到,最小值在离原点最近的地方取到稍等,马上给你上图,你就明白了再答:可行域是我涂黑的那部分可以轻松发现离原点最近和最
由约束条件x≥-1,y≥x,3x+2y≤5可求得1≥x≥-14x-3y+24=x+(3x-3y)+24当x取最大值即x=1时,(3x-3y)取最大值为0,即y=x时4x-3y+24有最大值即:4x-3
己知x,y为正实数,且xy=4x+y+12,有xy=4x+y+12>=2√(4xy)+12=4√(xy)+12,令t=√(xy)>0,有t^2-4t-12>=0,得t=6,所以xy的最小值为36
1.x^4+5x^2-6=(x^2+6)(x^2-1)=(x^2+6)(x-1)(x+1)2.(x+y)(x+y+2xy)+(xy+1)(xy-1)=(x+y)^2+2xy(x+y)+(xy+1)(x
(x+y-2xy)(x+y-2)+(1-xy)^2=(x+y)^2-2(1+xy)(x+y)+4xy+1-2xy+x^2y^2=(x+y)^2-2(1+xy)(x+y)+(1+xy)^2=(x+y)^
5x^2y/-1/2xy*3xy^2=-7.5x²y²
答:x+y=-1,xy=-2-5(x+y)+(x-y)+x(xy+y)=-5x-5y+x-y+xy(x+1)=-4x-6y+(-2)(x+1)=-4x-6y-2x-2=-6x-6y-2=-6(x+y)
(5-xy/x^3-x^2y)/(1-5/xy)^2*(y-x)^3=-(x-y)^3[-(xy-5)/x^2(x-y)]/[(xy-5)/xy]=(x-y)^3(xy-5)xy/x^2(x-y)(x
作图易知可行域为一个三角形,当直线z=3x+y过点A(3,-2)时,z最大是7,故答案为:7.
-5(x+y)+(x-y)+2(xy+y)=-5x-5y+x-y+2xy+2y=-4x-4y+2xy=-4(x+y)+2xy=-4×(-1)+2×(-2)=4+(-4)=0你有问题也可以在这里向我提问
再答:不明白的地方可以追问,望采纳,谢谢
因为xy/(x+y)=1/2所以x+y=2xy原式=3(x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1
1≤y=2x²-5x+4/(x-1)≤x,求解这个不等式,其中注意1≤x可以发现(x-2)²≤0x=2当x=2时,带入2x²-5x+4=y(x-1),得出y=2x+y=2
己知:x十y=3',Xy=一7,求:(1)x的平方十y平方的值,(=(x+y)²-2xy=9+14=23;2)(x一y)的平方的值=(x+y)²-4xy=9+28=37;很高兴为您
X=1,Y=2,XY=2;
x(x十y)(x一y)一x(x十y)²=x(x十y)[(x一y)一(x十y)]=x(x十y)*(-2y)=-2xy(x+y)=-2*6*5=-60
原式=4xy+8y+6x+2xy-5y-3x=6xy+3x+3y=6*3+3(x+y)=18-12=6