※ 引述《flygey (努力达成目标)》之铭言： : 1.Show that the parabola y=-x^2 and the line x-4y-18=0 intersect at right : angle at one of their points of intersection(於两个交点之一以直角相交) : (请问这题方程式如何设) : 2.Find the volume of the solid obtained by rotating the region bounded by : y=x-x^2 and y=0 about the line x=2 : (想请问这题界线是设 0-1 还是 0-2) : 3.The plane x+y+z=1 cuts the cylinder x^2+y^2=1 in an ellipse.Use Lagrange : multipliers to find the points on the ellipse that lie closest to and : farthest from the origin. : (这题方程式是 f(x,y,z)=x^2+y^2-1+入(x+y+z-1) 这样设吗 : 如果方程式求出xyz後要如何求最近最远点 ) : 请版友指导 : 感谢 3. Let f(x,y,z) = x^2 + y^2 + z^2 g(x,y,z) = x + y + z - 1 h(x,y,z) = x^2 + y^2 - 1 F(x,y,z) = f(x,y,z) + (λ)(g(x,y,z)) + (μ)(h(x,y,z)) = x^2 + y^2 + z^2 + (λ)(x + y + z - 1) + (μ)(x^2 + y^2 - 1) Fx = 2x + λ + 2μx = 0 ------(1) Fy = 2y + λ + 2μy = 0 ------(2) Fz = 2z + λ = 0 ------(3) (1) - (2) => 2x - 2y + 2μx - 2μy = 0 => 2x(1 + μ) - 2y(1 + μ) = 0 => 2(x - y)(1 + μ) = 0 x = y 或 μ = -1 当 x = y 时 x^2 + y^2 = 1 => x^2 + x^2 = 1 => (2)(x^2) = 1 1 => x^2 = --- 2 1 1 -1 -1 (x , y , z) = (----- , ----- , 1 - √2) , (----- , ----- , 1 + √2) √2 √2 √2 √2 当 μ = -1 时 代入(1)和(2)得 λ = 0 代入(3)得 2z = 0 => z = 0 x + y = 1 ------(4) x^2 + y^2 = 1 ------(5) (4) => y = 1 - x 代入(5)得 x^2 + (1 - x)^2 = 1 x^2 + x^2 - 2x + 1 = 1 (2)(x^2) - 2x = 0 => x^2 - x = 0 => (x)(x - 1) = 0 x = 0 或 1 (x , y , z) = (0 , 1 , 0)和(1 , 0 , 0) 1 1 1 1 f(---- , ---- , 1 - √2) = --- + --- + 3 - (2)(√2) = 4 - (2)(√2) √2 √2 2 2 -1 -1 1 1 f(---- , ---- , 1 + √2) = --- + --- + 3 + (2)(√2) = 4 + (2)(√2) √2 √2 2 2 f(0,1,0) = 1 f(1,0,0) = 1 -1 -1 当 (x,y,z) = (---- , ---- , 1 + √2) 时 √2 √2 有最远距离 √(4 + (2)(√2)) 当 (x,y,z) = (0 , 1 , 0) 和 (1 , 0 , 0) 时 有最短距离 1 --

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