Unformatted text preview: f is not deﬁned, so f must attain a min'
points are determined by: 3f _ 2 ﬁ
ﬁ — 23"— msyz _' 1
3f 2 Thus a: = :l:1 and y = :Ll, and it therefore follows that f has four critical points, , [—1,—1). Note that f has the value 3 for ‘ma. Thus, the points on the surface closest
to the point [0,0,0] are (1,1,1), [1,—1,—1), (~1, 1, ~1) and (—1, —1,1) and
the minimum distance is x/Z’; O 5 for Find the critical points of the functions in Exercises 1—8. 1 f($:y)=Izy2+xy+xy 2 flxiyl=r2+y2—xy+y 3. f(z,y}=:c2+y2+2xy—1IU 4. f(a:,y) =z2+§2+3my~3y 5 flay) =61+IL93 6. f(x,y)=xz—3xy+53—2y+6y2+8
7. f[z.y)=3x2+2xy+2x+yz+y+4
8 ﬁrst!) = sinlx2+ygl , a graph of the function z = (x3 — 3m)/(l + yz). . Refer to Figure 33.11, a graph of the function 2 = sin (sell/{1 + yg).
Where are the maximum and minimum points? 11. Minimize the distance to the origin from the plane a: — y+2z=3. 12. Find the distance from the plane given by rs + 25; + 32: — 10 = 0:
(a) To the origin. (b) To the point (1,1,1). l.”
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 Summer '10
 CarolThomas
 Math

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