将z看作是x和y的函数,即z=f(x,y)=x^4+y^4-4xy+1。\n\n首先求f(x,y)对x和y的一阶偏导数:\n∂f/∂x = 4x^3 - 4y\n∂f/∂y = 4y^3 - 4x\n\n然后解方程∂f/∂x = 0和∂f/∂y = 0,求出极值点。\n\n令4x^3 - 4y = 0,得到x^3 = y\n将y代入∂f/∂y = 0,得到4(x^3)^3 - 4x = 0\n化简得到4x^9 - 4x = 0\n解得x = 0或x = ±1\n\n当x = 0时,由x^3 = y,得到y = 0\n当x = 1时,由x^3 = y,得到y = 1\n当x = -1时,由x^3 = y,得到y = -1\n\n所以极值点为(0, 0),(1, 1),(-1, -1)。\n\n接下来求二阶偏导数:\n∂²f/∂x² = 12x^2\n∂²f/∂y² = 12y^2\n∂²f/∂x∂y = -4\n对于(0, 0),有∂²f/∂x² = 0,∂²f/∂y² = 0,∂²f/∂x∂y = -4\n构造二阶偏导数判别式D = (∂²f/∂x²)(∂²f/∂y²) - (∂²f/∂x∂y)² = 0 - (-4)² = 16 > 0,且∂²f/∂x² = 0 < 0,所以(0, 0)是极小值点。\n\n对于(1, 1),有∂²f/∂x² = 12,∂²f/∂y² = 12,∂²f/∂x∂y = -4\n构造二阶偏导数判别式D = (∂²f/∂x²)(∂²f/∂y²) - (∂²f/∂x∂y)² = (12)(12) - (-4)² = 144 - 16 = 128 > 0,且∂²f/∂x² = 12 > 0,所以(1, 1)是极大值点。\n\n对于(-1, -1),有∂²f/∂x² = 12,∂²f/∂y² = 12,∂²f/∂x∂y = -4\n构造二阶偏导数判别式D = (∂²f/∂x²)(∂²f/∂y²) - (∂²f/∂x∂y)² = (12)(12) - (-4)² = 144 - 16 = 128 > 0,且∂²f/∂x² = 12 > 0,所以(-1, -1)是极大值点。\n\n综上,z=x^4+y^4-4xy+1的极小值点为(0, 0),极大值点为(1, 1)和(-1, -1)。

求函数z=x^4+y^4-4xy+1的极值点和极值

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