x,y满足x²/6+y²/2=1,求z=x-y的最大值和最小值,用参数方程怎么解?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/09 02:16:02
![x,y满足x²/6+y²/2=1,求z=x-y的最大值和最小值,用参数方程怎么解?](/uploads/image/z/4525291-19-1.jpg?t=x%2Cy%E6%BB%A1%E8%B6%B3x%26%23178%3B%2F6%2By%26%23178%3B%2F2%3D1%2C%E6%B1%82z%3Dx-y%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC%2C%E7%94%A8%E5%8F%82%E6%95%B0%E6%96%B9%E7%A8%8B%E6%80%8E%E4%B9%88%E8%A7%A3%3F)
x)Щ|{m+Ԕ
-ʹ+,#[Cgl+t+jy6O<bouOY
Ϧ|YCߓ//I*ҧv64w=6C2O;fV>e_|nX7*̃u6<ٽVW.ס Vbk Jo7D
w r0E!m@! 3G
x,y满足x²/6+y²/2=1,求z=x-y的最大值和最小值,用参数方程怎么解?
x,y满足x²/6+y²/2=1,求z=x-y的最大值和最小值,用参数方程怎么解?
x,y满足x²/6+y²/2=1,求z=x-y的最大值和最小值,用参数方程怎么解?
x²/6+y²/2=1
则x=√6cosθ
y=√2sinθ
所以z=-(√2sinθ-√6cosθ)
=-2√2sin(θ-π/3)
所以最大值是2√2
最小值是-2√2