《高中数学 第三章 导数应用 3.2导数在实际问题中的应用 3.2.2.2 导数在实际问题中的应用课件 北师大版选修22》由会员分享，可在线阅读，更多相关《高中数学 第三章 导数应用 3.2导数在实际问题中的应用 3.2.2.2 导数在实际问题中的应用课件 北师大版选修22（28页珍藏版）》请在金锄头文库上搜索。
1、第2 2课时导数在实际问题中的应用ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1.了解导数在解
2、决实际问题中的作用.2.掌握利用导数解决简单的实际生活中的优化问题.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODA
3、OHANG目标导航1.解决实际问题的关键在于建立数学模型和目标函数,把“问题情境”译为“数学语言”,找出问题的主要关系,抽象成数学问题,然后用可导函数求最值的方法求最值.2.解决优化问题的基本思路.上述解决优化问题的过程是一个典型的数学建模过程.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITAN
4、GYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航解析:由题设知y=x2-39x-40,所以当x=40时,y取得极小值,也是最小值.故为使耗电量最小,其速度应定为40.答案:40ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练D
5、IANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四【例1】为了在夏季降温和冬季供暖时减少能源损耗,房屋的屋顶和外墙需要建造隔热层.某幢建筑物要建造可使用20年的隔热层,每厘米厚的隔热层建造成本为6万元.该建筑物每年的能源消耗费用C(单位:万元)与隔热层厚度x(单位:cm)满足关系:C(x)= (0x10).若不建隔热层,每年能源消耗费用为8万元.设f
6、(x)为隔热层建造费用与20年的能源消耗费用之和.(1)求k的值及f(x)的表达式;(2)当隔热层修建多厚时,总费用f(x)达到最小,并求最小值.分析:根据题设条件构造函数关系,再应用导数求最值.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例
7、透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI
8、SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四反思反思解决应用问题时,步骤“设、列、解”缺一不可,写出函数关系式及定义域后用导数求最值,解决最值问题.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITA
9、NGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四【变式训练1】一艘轮船在航行时每小时的燃料费和它的速度的立方成正比.已知速度为每小时10海里时,燃料费是每小时6元,而其他与速度无关的费用是每小时96元,问轮船的速度是多少时,航行1海里所需的费用总和最小?ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练D
10、IANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透
11、析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四【例2】某厂生产某种电子元件,如果生产出一件正品,那么可获利200元,如果生产出一件次品,那么损失100元.已知该厂制造电子元件过程中,次品率p与日产量x的函数关系是 (xN+),(1)将该厂的日盈利额T(单位:元)表示为日
12、产量x(单位:件)的函数;(2)为获最大盈利,该厂的日产量应定为多少件?分析:根据次品率,先计算出正品数与次品数,再用获利总数减去损失总数可得盈利.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SU
13、ITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随
14、堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四反思反思解决此类有关利润的实际应用题,应灵活运用题设条件,建立利润的函数关系,常见的基本等量关系有:(1)利润=收入-成本;(2)利润=每件产品的利润销售件数.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理S
15、UITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN
16、随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOU
17、XI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四【例3】如图,要设计一张矩形广告,该广告含有大小相等的左右两个矩形栏目(即图中阴影部分),这两栏的面积之和为18 000 cm2,四周空白的宽度为10 cm,两栏之间的中缝空白的宽度为5 cm.怎样确定广告的高与宽的尺寸(单位:cm),能使矩形广告的面积最小?分析:设出适当的变量,把广告面积用该变量表示出来,然后用导数解答最值问题.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI
18、典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISH
19、ISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四令S0,得x140;令S0,得20x140.函数在(140,+)上是增加的,在(20,140)上是减少的,S(x)的最小值为S(140).当
20、x=140时,y=175,即当x=140,y=175时,S取得最小值24 500 cm2.故当广告的高为140 cm,宽为175 cm时,可使广告的面积最小.反思反思解决面积、容积的最值问题,要正确引入变量,将面积或容积表示为变量的函数,结合实际问题的定义域,利用导数求解函数的最值.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZH
21、ISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四【变式训练3】用长为90 cm,宽为48 cm的长方形铁皮做一个无盖的容器,先在四个角分别截去一个小正方形,再把四边翻转90,然后焊接成如图所示的容器,问该容器的高为多少时,容器的容积最大?最大容积是多少?ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理S
22、UITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四解:设容器的高为x cm,容器的容积为V(x)cm3.则V(x)=x(90-2x)(48-2x)=4x3-276x2+4 320x(0x24).V(x)=12x2
23、-552x+4 320=12(x2-46x+360)=12(x-10)(x-36)(0x24).令V(x)=0,得x1=10,x2=36(舍去).当0x0,V(x)是增加的;当10x24时,V(x)0,V(x)是减少的.因此,在定义域(0,24)内,函数V(x)只有在x=10处取得最大值,其最大值为V(10)=10(90-20)(48-20)=19 600.故当容器的高为10 cm时,容器的容积最大,最大容积是19 600 cm3.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随
24、堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四易错点:应用函数模型时,不注意实际问题的定义域而致错【例4】现有一批货物由海上从A地运往B地,已知轮船的最大航行速度为35海里/时,A地至B地之间的航行距离约为500海里,每小时的运输成本由
25、燃料费和其余费用组成,轮船每小时的燃料费与轮船速度的平方成正比(比例系数为0.6),其余费用为每小时960元.(1)把全程运输成本y(单位:元)表示为速度x(单位:海里/时)的函数.(2)为了使全程运输成本最小,轮船应以多大速度行驶?ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLI
26、AN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLIT
27、OUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航Z
28、HISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理S
29、UITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演
30、练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 53已知某生产厂家的年利润y(单位:万元)与年产量x(单位:万件)之间的函数关系式为y=- x3+81x-234,则使该生产厂家获取最大年利润的年产量为()A.13万件B.11万件C.9万件D.7万件解析:y=-x2+81,令y=0,得x=9(x=-9舍去),且经讨论知x=9是函数的极大值点,也是最大值点,故厂家获得最大年利润的年产量是9万件.答案:CZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN
31、随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典
32、例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5
