高専の数学解説

第7章 7.17

(1)

z=x(1−x2−y2)=x−x3−xy2z = x(1-x^2-y^2) = x - x^3 - xy^2

zx=1−3x2−y2zy=−2xyzxx=−6xzyy=−2xzxy=−2yz_x = 1-3x^2-y^2 \qquad z_y = -2xy \qquad z_{xx} = -6x \qquad z_{yy} = -2x \qquad z_{xy} = -2y

zx=zy=0z_x = z_y = 0 のとき (x,y)=(0,±1),(±13,0)(x,y) = (0,\pm1), (\pm\dfrac{1}{\sqrt{3}}, 0)

H(x,y)=12x2−4y2H_{(x,y)} = 12x^2 - 4y^2

(0,±1)(0,\pm1)のとき H<0H<0

(±13,0)(\pm\dfrac{1}{\sqrt{3}},0)のとき H>0H>0

zxx(13,0)<0z_{xx(\frac{1}{\sqrt{3}},0)} < 0 極大 zxx(−13,0)>0z_{xx(-\frac{1}{\sqrt{3}},0)} > 0 極小

よって(13,0)(\dfrac{1}{\sqrt{3}},0)のとき極大値233\dfrac{2}{3\sqrt{3}} (−13,0)(-\dfrac{1}{\sqrt{3}},0)のとき極大値−233-\dfrac{2}{3\sqrt{3}}

(2)

z=(x2+y2)2−2(x2−y2)z = (x^2+y^2)^2 - 2(x^2-y^2)

zx=4x(x2+y2)−4x=4x(x2+y2−1)z_x = 4x(x^2+y^2)-4x = 4x(x^2+y^2-1)

zy=4y(x2+y2)+4y=4y(x2+y2+1)z_y = 4y(x^2+y^2)+4y = 4y(x^2+y^2+1)

zxx=4(x2+y2−1)+8x2=4(3x2+y2−1)z_{xx} = 4(x^2+y^2-1)+8x^2 = 4(3x^2+y^2-1)

zyy=4(x2+y2+1)+8y2=4(x2+3y2+1)z_{yy} = 4(x^2+y^2+1)+8y^2 = 4(x^2+3y^2+1)

zxy=8xyz_{xy} = 8xy

zx=zy=0z_x = z_y = 0のとき (x,y)=(0,0),(±1,0)(x,y) = (0,0), (\pm1,0)

H(x,y)=4(3x2+y2−1)⋅4(x2+3y2+1)−64x2y2H_{(x,y)} = 4(3x^2+y^2-1)\cdot 4(x^2+3y^2+1) - 64x^2y^2

H(0,0)=−16<0H_{(0,0)} = -16 < 0

H(±1,0)=64>0H_{(\pm1,0)} = 64 > 0

zxx(±1,0)=8>0z_{xx(\pm1,0)} = 8 > 0 極小

よって(±1,0)(\pm1,0)のとき極小値−1-1

(3)

z=x4+y2+2x2−4xy+1z = x^4 + y^2 + 2x^2 - 4xy + 1

zx=4x3+4x−4yzy=2y−4xzxx=12x2+4zyy=2zxy=−4z_x = 4x^3 + 4x - 4y \qquad z_y = 2y - 4x \qquad z_{xx} = 12x^2 + 4 \qquad z_{yy} = 2 \qquad z_{xy} = -4

zx=zy=0z_x = z_y = 0 のとき (x,y)=(0,0),(±1,±2)(x,y) = (0,0),(\pm1,\pm2)

H(x,y)=2(12x2+4)−16H_{(x,y)} = 2(12x^2+4) - 16

(0,0)(0,0) のとき H(0,0)<0H_{(0,0)} < 0

(±1,±2)(\pm1,\pm2) のとき H(±1,±2)>0H_{(\pm1,\pm2)} > 0

zxx(±1,±2)>0z_{xx(\pm1,\pm2)} > 0 極小

よって (±1,±2)(\pm1,\pm2) のとき極小値 00 （複合同順）

(4)

z=(x2+2y2)e−x2−y2z = (x^2+2y^2)e^{-x^2-y^2}

zx=2xe−x2−y2−2x(x2+2y2)e−x2−y2=2xe−x2−y2(1−x2−2y2)z_x = 2xe^{-x^2-y^2} - 2x(x^2+2y^2)e^{-x^2-y^2} = 2xe^{-x^2-y^2}(1-x^2-2y^2)

zy=4ye−x2−y2−2y(x2+2y2)e−x2−y2=2ye−x2−y2(2−x2−2y2)z_y = 4ye^{-x^2-y^2} - 2y(x^2+2y^2)e^{-x^2-y^2} = 2ye^{-x^2-y^2}(2-x^2-2y^2)

zxx=2e−x2−y2(1−x2−2y2)−4x2e−x2−y2(1−x2−2y2)−4x2e−x2−y2z_{xx} = 2e^{-x^2-y^2}(1-x^2-2y^2) - 4x^2e^{-x^2-y^2}(1-x^2-2y^2) - 4x^2e^{-x^2-y^2}

zyy=2e−x2−y2(2−x2−2y2)−4y2e−x2−y2(2−x2−2y2)−8y2e−x2−y2z_{yy} = 2e^{-x^2-y^2}(2-x^2-2y^2) - 4y^2e^{-x^2-y^2}(2-x^2-2y^2) - 8y^2e^{-x^2-y^2}

zxy=−4xye−x2−y2(1−x2−2y2)−8xye−x2−y2z_{xy} = -4xye^{-x^2-y^2}(1-x^2-2y^2) - 8xye^{-x^2-y^2}

zx=zy=0z_x = z_y = 0 のとき (x,y)=(0,0),(0,±1),(±1,0)(x,y) = (0,0),(0,\pm1),(\pm1,0)

H(0,0)=4>0H_{(0,0)} = 4 > 0 zxx(0,0)>0z_{xx(0,0)} > 0 極小

H(0,±1)>0H_{(0,\pm1)} > 0 zxx(0,±1)<0z_{xx(0,\pm1)} < 0 極大

H(±1,0)<0H_{(\pm1,0)} < 0

よって (0,0)(0,0) のとき極小値 00 (0,±1)(0,\pm1) のとき極小値 2e\dfrac{2}{e}

(5)

z=xe−x2−y2z = xe^{-x^2-y^2}

zx=e−x2−y2−2x2e−x2−y2=e−x2−y2(1−2x2)zy=−2xye−x2−y2z_x = e^{-x^2-y^2} - 2x^2e^{-x^2-y^2} = e^{-x^2-y^2}(1-2x^2) \qquad z_y = -2xye^{-x^2-y^2}

zxx=−4xe−x2−y2+2x(2x2−1)e−x2−y2=2xe−x2−y2(2x2−3)z_{xx} = -4xe^{-x^2-y^2} + 2x(2x^2-1)e^{-x^2-y^2} = 2xe^{-x^2-y^2}(2x^2-3)

zyy=−2xe−x2−y2+4xy2e−x2−y2=2xe−x2−y2(2y2−1)z_{yy} = -2xe^{-x^2-y^2} + 4xy^2e^{-x^2-y^2} = 2xe^{-x^2-y^2}(2y^2-1)

zxy=2y(2x2−1)e−x2−y2z_{xy} = 2y(2x^2-1)e^{-x^2-y^2}

zx=zy=0z_x = z_y = 0 のとき (x,y)=(±12,0)\qquad (x,y) = (\pm\dfrac{1}{\sqrt2},0)

H(12,0)>0H_{(\frac{1}{\sqrt2},0)} > 0

H(−12,0)>0H_{(-\frac{1}{\sqrt2},0)} > 0

zxx(12,0)=−12e−12=−12e<0極大z_{xx(\frac{1}{\sqrt2},0)} = -\frac{1}{\sqrt2}e^{-\frac12} = -\frac{1}{\sqrt{2e}} < 0 \quad \text{極大}

zxx(−12,0)=12e>0極小z_{xx(-\frac{1}{\sqrt2},0)} = \frac{1}{\sqrt{2e}} > 0 \quad \text{極小}

よって (12,0)(\dfrac{1}{\sqrt2},0) のとき極大値 12e\dfrac{1}{\sqrt{2e}} \qquad (−12,0)(-\dfrac{1}{\sqrt2},0) のとき極小値 −12e-\dfrac{1}{\sqrt{2e}}

解説: r31bn1z