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Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

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_Answers_441<br />

e=3 1^25 (x\)* (f/-f2) 2 and consequently exist only <strong>in</strong>side and on<br />

the boundary of thecircle (x I) 2<br />

-|- (l/ + 2> 2 = 25, at the po<strong>in</strong>ts of which both<br />

functions assume the value 2 = 3. This value is the least for the first function<br />

and is the greatest for the second. 2020. One of the functions def<strong>in</strong>ed by the<br />

equation has a maximum (*max"-~~~ 2) for x 1, r/ = 2, the other has a<br />

m<strong>in</strong>imum (2m<strong>in</strong> = 1) for x = 1, y 2, both functions have a boundary extremum<br />

at the po<strong>in</strong>ts of the curve 4jt s<br />

2<br />

4i/<br />

12* + 16r/ 33=0. 2021. 2 max =-j for<br />

*=#=-. 2022. 2max = 5 for x=l, = f/ 2; 2m<strong>in</strong> ---- 5 for x = 1, #--2<br />

2023. imta for * = , ^~. 2024. ~. --<br />

. max<br />

j/ = 5<br />

+ *:i, 2025.<br />

= 9 for x = 1, = j/ 2, 2= 2, "max^ 9 f r *=1, = r/ 2, 2 = 2.<br />

2026. MMIX -=fl f r *= fl = I/ 2 = 0; Wm<strong>in</strong>^C for x = t/=0 2^C.<br />

2027. w max = 2.4 2 .6Mor x = 2, y=4, 2=6. 2028. u max -4 4 / 27 at the po<strong>in</strong>ts<br />

(f f I)' (4- f T)= (f f T)'---<br />

2, 1) (2, 1, 2) (1, 2, 2). 2030. a) Greatest value 2 = 3 for x = 0, y=l;<br />

b) smallest value 2 = 2 for x== 1, j/ = 0. 2031. a) Greatest value 2= , for<br />

/^<br />

/"T 2 /""2"<br />

TFT for x== ^ V j<br />

"3 ^^ T 3 ; smallest value z== "~<br />

y I/ -IT; b) greatest value 2 = 1 for *= 1, y-^0; smallest value<br />

z = \ for x = 0, t/= 1. 2032. Greatest value 2 = ^<br />

for x = t/ = Y (<strong>in</strong>-<br />

ternal maximum); smallest value 2 = for x y-~Q (boundary m<strong>in</strong>imum).<br />

2033. Greatest value 2 = 13 for x = 2, y = 1 (boundary maximum); smallest<br />

value ? = 2 for x y = \ (<strong>in</strong>ternal m<strong>in</strong>imum) and for x = 0, y = 1 (boun-<br />

dary m<strong>in</strong>imum). 2034. Cube. 2035. J/2V, j/2V, -- ^2V* 2036. Isosceles<br />

triangle. 2037. Cube. 2038. a = \/a \/a ><br />

2<br />

2040. Sides of the triangle are ~p,~/>, and .<br />

fe 44<br />

\/a >/a.<br />

2041. Jgaa<br />

2039. Ai f<br />

v<br />

ll 1a ,<br />

i><br />

m, -i- m 2 -f m,<br />

2Q42 + a + b<br />

s=s3i<br />

c<br />

2Q43. The dimensions of the<br />

parallelepiped<br />

*2a 2b 2c<br />

are , -rp 7^-* "T^' where a, b, and c are the semi-<br />

Y 3 V 3 V 3<br />

axes of the ellipsoid. 2044. * = j/ = 26+2V, * = -- 2045. x= - ,<br />

y ~T 2046 - Major axis, 2a = 6, m<strong>in</strong>or axis, 26 = 2. H<strong>in</strong>t. The square of<br />

the distance of the po<strong>in</strong>t (x,y) of the ellipse from its centre (coord<strong>in</strong>ate orig<strong>in</strong>)<br />

2 2<br />

is equal to x . -f-r/ The problem reduces to f<strong>in</strong>d<strong>in</strong>g the extremum of the function<br />

x*-\-y* provided 5*2<br />

2<br />

+ Bxy + = 5t/ 9. 2047. The radius of the base of the cyl<strong>in</strong>der<br />

^ ,<br />

-|- J<br />

.

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