f(x,y)=x^2/y 判断凹凸性
时间: 2024-04-27 12:22:00 浏览: 180
判断多边形点的凹凸性
我们可以通过计算二阶偏导数来判断$f(x,y)$的凹凸性。具体地,如果$f(x,y)$的二阶偏导数满足以下条件:
1. 当$\frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2 > 0$时,$f(x,y)$为双曲抛物面,即为非凹非凸函数;
2. 当$\frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2 < 0$时,$f(x,y)$为椭圆抛物面,即为严格凸函数;
3. 当$\frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2 = 0$时,无法判断凹凸性。
我们来计算一下$f(x,y)$的二阶偏导数:
$$
\frac{\partial^2 f}{\partial x^2} = 2/y,\quad \frac{\partial^2 f}{\partial y^2} = -2x^2/y^3,\quad \frac{\partial^2 f}{\partial x\partial y} = 0
$$
将其代入公式中,得到:
$$
\frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2 = -4/x^2y^4
$$
因为上式小于零,所以$f(x,y)$为严格凸函数。
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