functionvalues = f[x, y, z] /. points
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10.8 Find all relative extrema of the function f(x, y) = x2 y2. Sketch the surface.
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f[x_, y_] = x2 y2; Solve[{ xf[x, y] 0, yf[x, y] 0}, {x, y}] {{x 0, y 0}} 2 d[x_, y_]= {x, 2} f[x, y] (y, 2} f[x, y] ( x, y f[x, y]); d[0, 0] 4
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Negative number; saddle point at (0, 0).
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Multivariate Calculus
10.9 Find all relative extrema of the function f ( x , y) = xye x
SOLUTION
2 y2
. Sketch the surface.
f[x_, y_] = x y Exp[ x2 y2] pdx = x f[x, y] //Factor x x
2 y2
( 1 + 2 x2)y x( 1 + 2 y2)
pdy = y f[x, y] //Factor
2 y2
If we try to use Solve to find where the partial derivatives are 0, we will get an error message due 2 2 to the presence of the (non-algebraic) exponential. However, since x y cannot equal zero, we can ignore its presence.
Solve[{ 1 + 2 x2) y 0, x ( 1 + 2 y2) 0, {x, y}]
x 0, y 0}, x 1 , y 1 , x 1 , y 1 , 2 2 2 2 x 1 ,y 1 , x 1 ,y 1 2 2 2 2
2 d[x_, y_] = {x,2} f[x, y] (y,2} f[x, y] ( x,y f[x, y]);
d[0, 0] 1 d[ 1 / 2, 1 / 2] 4 2 {x,2} f[x, y]/.{ x 1 / 2, y 1 / 2 } 2 d[ 1 / 2, 1 / 2] 4 2 {x,2} f[x, y]/.{ x 1 / 2, y 1 / 2 } 2 d[1 / 2, 1 / 2] 4 2 {x,2} f[x, y]/.{ x 1 / 2, y 1 / 2 } 2 d[1/ 2, 1/ 2] 4 2 {x,2} f[x, y]/.{ x 1 / 2, y 1 / 2 } 2
Relative maximum at Relative minimum at Relative minimum at Relative maximum at Negative number; no relative extremum.
1 1 , . 2 2
1 1 , . 2 2
1 1 , . 2 2
1 1 , . 2 2
Multivariate Calculus
We sketch the surface showing two views.
Plot3D[f[x, y], {x, 3, 3}, {y, 3, 3}, PlotPoints 30, ViewPoint {1.391, 3.001, 0.713}, PlotRange All] Plot3D[f[x, y], {x, 3, 3}, {y, 3, 3}, PlotPoints 30, ViewPoint {0.617, 3.318, 0.245}, PlotRange All]
2 2 0.1 0.0 0.1 2 0 2 2 0 2 0 2
2 0.1 0.0 0.1
10.10 Use Lagrange multipliers to find the points on the circle x2 + y2 2x 4y = 0 closest to and farthest from P(4, 4).
SOLUTION
circle = ContourPlot[x2 + y2 2 x 4 y 0, {x, 5, 5 }, {y, 1, 5}]; point = Graphics[{PointSize[.01], Point[{4, 4}]}]; Show[circle, point, Axes True, Frame False, AspectRatio Automatic]
5 4 3 2 1 4 2 1 2 4
f[x_, y_] = (x 4)2 + (y 4)2; g[x_, y_] = x2 + y2 2 x 4 y;
We minimize the square of the distance from P.
conditions = Eliminate[{ xf[x, y] k x g[x, y], yf[x, y] k y g[x, y], g[x, y] 0}, k] 2 x 4 + 3 y & & 52 y + 13 y 2 32 points = Solve[conditions]
x 1 (13 3 65 ), y 2 (13 65 ) , x 1 (13 + 3 65 ), y 2 (13 + 65 ) 13 13 13 13 Based upon the computed distances, the first point is farthest from P and the second point is closest.
f[x,y] /. points //N {5.84162, 1.36948}
Multivariate Calculus
10.11 Find the points on the sphere x2 + y2 + z2 = 1 that are closest to and farthest from (1, 2, 3).
SOLUTION
f[x_, y_, z_] = (x 1)2 + (y 2)2 + (z 3)2; g[x_, y_, z_] = x2 + y2 + z2 1; conditions = Eliminate[{ x f[x, y,z] x g[x, y,z], y f[x, y,z] k y g[x, y,z], z f[x, y,z] k z g[x, y,z], g[x, y,z] 0}, k]