二、解：两边同时对x求偏导得：ex?ysin?x?z??ex?ycos?x?z??1????z???0 ?x?故sin?x?z??z???1??tan?x?z??1 ?xcos?x?z???z?同理，两边同时对y求偏导得：ex?ysin?x?z??ex?ycos?x?z????0
??y?故sin?x?z??z????tan?x?z? ?ycos?x?z??z?z?zx?2y?x?2,故? ?y?y?y2z?22y?2z三、解：两边同时对y求偏导得：四、解：?dz?F'?zF2'zF'?F'?z?z?z?zdx?dy 而，??1??12 ?x?y?xyF1'?xF2'?yyF1'?xF2'故dz??(F1'?zF2')dx?(zF1'?F2')dy
yF1'?xF2'?z?xyz ze?xy五、解:两边同时对x求偏导得：??2z?2?xy?zz??z?e?xy??yz?ez?y??z?yez?xy2?yzez??y2z??x??x???x22zze?xye?xy?????yz?yez?xy2?yzez??y2z?ez?xy??ez?xy?3?2y2zez?2xy3z?y2z2ez?ez?xy?3
dyy?yexy?xy. 8*解：
dxxe?x又?ex'?dz(x?z)exsin(x?z)?dz??x?zsint?x???dt??e??1???dx?1?sin(x?z) 0tx?z?dx???'xyx??duy?ye(x?z)e?f1??f2?xy?f3??1?? dxxe?x?sin(x?z)?习题十七
一、（1）B，（2）B，（3）C
2xx?y2?2y?e2x，fy?x,y???2y?2?e2x， 二、解：fx?x,y??2e???1?fy?x,y??0令? ，解出得唯一驻点(,?1).
2??fy?x,y??0又A?fxx?1??,?1??2?2x22x??1??4ex?y?2y?4e??????,?1??2??2e，
B?fxy?1??,?1??2?2x???(4y?4)e???1??,?1??2?''?0， C?fyy?1??,?1??2??2ex2?1??,?1??2??2e，
e?1?故AC?B2?4e2?0且A?0,有极小值f?,?1???.
2?2?三、解：设容积为V，长、宽、高分别为x,y,V, xy则：S?x,y??2xy?2V2V2V2V?,令Sx?x,y??2y?2?0,Sy?x,y??2x?2?0 yxxy解之得唯一驻点：x?y?3V， 又Sxx?x,y?Sxy?x,y??3V,3V?3?Syy?x,y??3V,3V??A?C?4,
?3V,V??B?2,故AC?B2?0且A?0,有极小值，
即长、宽、高分别为3V时材料最少．22四、解：（1）利润函数L?R?(x1?x2)?15?13x1?31x2?8x1x2?2x1?10x2
??Lx1??4x1?8x2?13?0令?解得唯一驻点（0.75，1.25），
L??8x?20x?31?0?x12?2?2L?2L?2LB???8又A?2??4， ，C???20，
?x1x2?x1?x22AC-B2?16?0,A?0，故点（0.75，1.25）为极大值点.
即此时最优广告策略为：用0.75万元作电台广告，用1.25万元作报纸广告. （2）令拉格朗日函数F(x1,x2,?)?L(x1,x2)??(x1?x2?1.5)
?15?13x1?31x2?8x1x2?2x12?10x22??(x1?x2?1.5)
Fx1?13?8x2?4x1???0???x?0令Fx2?31?8x1?20x2???0???1，
??x2?1.5F??x1?x2?1.5?0?即广告费用1.5万元全部用于报纸广告，可使利润最大.
习题十八
?y2?4x一. 解 定义域为? 220?y?x?1?1x?,y?02limf(x,y)?2 3ln4二.解 f(1,x)?yy2xy2
x?y三 解
?z?z?ex(sinxy?ycosxy),?exxsinxy, dz?ex(sinxy?ycosxy)dx?exxsinxydy. ?x?y四 解
2y2y?z1?z1?,?,dz?dx?dy.。 2222?xx?y?yx?yx?yx?y?dx
dz1?dx?dy，dzx?y?12x?1,y?0五 解 f(xy,f(x,y))?3xy?2f(x,y)?3xy?2(3x?2y)。
x2(1?y)六解 f(x,y)?
1?y?z2x?y?zxy2?2y(x2?xy)2x?y2x2y?xy七解 ?,?, dz?dx?dy。 2423?xy?yyyy八解
?z?z?f1?exsiny?f2?2x, ?f1?excosy?f2?2y ?x?ydz?(f1?exsiny?f2?2x)dx?(f1?excosy?f2?2y)dy.
九解
?z11??2f(xy)?f?(xy)y?y??(x?y)， ?xxx?z1?f?(xy)x??(x?y)?y??(x?y) ?yxdz?[?1yf(xy)?f?(xy)?y??(x?y)]dx?[f?(xy)??(x?y)?y??(x?y)]dy 2xx22?f?2f?f?x2y2?x2y2?e?y，?e?x，2?e?xy?(?2xy3)， 十解 ?x?x?y22?2f?2f?x2y222?e?(1?2xy)，2?e?xy?(?2x3y)， ?x?y?yx?2f?2fy?2f?x2y2222222?x2y2故 ?2??e?[?2xy?2?4xy?2xy]??2ey?x2?x?yx?y2du?f?fdy?fdz?f?fyexy?fz十一解 。?????()?() xyzdxdx?ydx?zdxdx?y1?xe?ze?xz?y?x?z?z1?e?xez?十二解 dz?dx?dy??y?x?x?y1?xzey?xyx?yx1?xe?z?1?e?z(1?)x d?xd?yd?xz?y?x?z?yx1?xe1?xedy十三、解：（1）设利润函数为L，则
22L?PQ11?PQ22?2(Q1?Q2)?5?16Q1?10Q2?2Q1?Q2?5，
LQ1?16?4Q1?0???Q1?4令 ???LQ2?10?2Q2?0?Q?5??2又因最大利润一定存在，必在驻点处取得，故当Q1?4,P1?10,Q2?5,P2?7时有最大利润
L?52.
（2）令P1?P2?P，那么18?2Q1?P?Q1?9?P，12?Q2?P?Q2?12?P 232则L?P(Q1?Q2)?2(Q1?Q2)?5?24P?P?47
2dL?24?3P?0，得唯一驻点P?8. 令dP又因最大利润一定存在，必在驻点处取得，故当P有最大利润L?49.1?P2?8,Q1?5,Q2?4时，
显然实行价格差别策略时总利润要大些.
习题十九
一、(1).C (2).C (3).B (4).D (5).D (6).D (7).A (8).C (9).A 二、（1）> (2)
??D1?x?yd? (3)?dx? 022 1 x2 0f(x,y)dy (4)
3? 2三 1 解：
822。 (x?y)d??((x?y)dy)dx?????3D?1?122111x1?siny?(siny)(siny)d???dx?dy???(y?y2)?dy?1?sin1 2解：??yyy?D0x0?elnxx?ye3解：
??eDDx?yd???dx?e102dy??ex(x?1)dx?ee(e?2)
122?x4解：
2(3x?2y)d??dx(3x?2y)dy?(3xy?y)?????0002?x0dx
?12?y? 2 0(6x?3x2?(x?2)2)dx?120 3 5解：
??xyd???dx(D0y2?111xydx)dy??((2?y)2y?y3)dy?.
22301x四 1解：
??xyd???xdx?D0x2111yydy??ydy?xdx??y(x22)dy??.
2511y00y2221y1y1139 2解：??d???dx?ydy???3x2dx??xdx?.
xxx2x24D111 3解：
22x??(2x?y)d???dy?1(2x?y)dx?D 1 y14?2y 2 y19. 6xx55. 4解：??(2?y?)d???dy?(2?y?)dx?222D?22y2bd12b1d2 5解：
??f(x)f(y)d???dx?f(x)f(y)dy??f(x)dx?f(y)dy).
12Dacac12x 6解：原式=dx0??f(x,y)dy.
x21?22acos?七 1解：原式=
?d?0?03r3dr??a4.
41?31cos?2解：原式=
1?4?d??01dr?ln2?31?2.
1?2Rcos?八 1解：原式=
1??2?d??0R2?r2rdr??3R3.
1?2 2解：原式=
?01d??ln(1?r2)rdr??(2ln2?1).
20习题二十
1
3212522一 解： ?dy?xydx??y(1?y)dy?(22?1)
30150?1?y2211?y2