《高中數(shù)學(xué) 第三章 導(dǎo)數(shù)應(yīng)用 3.2導(dǎo)數(shù)在實際問題中的應(yīng)用 3.2.2.2 導(dǎo)數(shù)在實際問題中的應(yīng)用課件 北師大版選修22》由會員分享，可在線閱讀，更多相關(guān)《高中數(shù)學(xué) 第三章 導(dǎo)數(shù)應(yīng)用 3.2導(dǎo)數(shù)在實際問題中的應(yīng)用 3.2.2.2 導(dǎo)數(shù)在實際問題中的應(yīng)用課件 北師大版選修22（28頁珍藏版）》請在裝配圖網(wǎng)上搜索。

1、第2 2課時導(dǎo)數(shù)在實際問題中的應(yīng)用MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1.了解導(dǎo)數(shù)在解

2、決實際問題中的作用.2.掌握利用導(dǎo)數(shù)解決簡單的實際生活中的優(yōu)化問題.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI

3、SHULI知識梳理1.解決實際問題的關(guān)鍵在于建立數(shù)學(xué)模型和目標(biāo)函數(shù),把“問題情境”譯為“數(shù)學(xué)語言”,找出問題的主要關(guān)系,抽象成數(shù)學(xué)問題,然后用可導(dǎo)函數(shù)求最值的方法求最值.2.解決優(yōu)化問題的基本思路.上述解決優(yōu)化問題的過程是一個典型的數(shù)學(xué)建模過程.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DI

4、ANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理解析:由題設(shè)知y=x2-39x-40,所以當(dāng)x=40時,y取得極小值,也是最小值.故為使耗電量最小,其速度應(yīng)定為40.答案:40MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析

5、SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四【例1】 為了在夏季降溫和冬季供暖時減少能源損耗,房屋的屋頂和外墻需要建造隔熱層.某幢建筑物要建造可使用20年的隔熱層,每厘米厚的隔熱層建造成本為6萬元.該建筑物每年的能源消耗費用C(單位:萬元)與隔熱層厚度x(單位:cm)滿足關(guān)系:C(x)= (0 x10).若不建隔熱層,每年能源消耗費用為8萬元.

6、設(shè)f(x)為隔熱層建造費用與20年的能源消耗費用之和.(1)求k的值及f(x)的表達(dá)式;(2)當(dāng)隔熱層修建多厚時,總費用f(x)達(dá)到最小,并求最小值.分析:根據(jù)題設(shè)條件構(gòu)造函數(shù)關(guān)系,再應(yīng)用導(dǎo)數(shù)求最值.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYAN

7、LIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISH

8、ISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四反思反思解決應(yīng)用問題時,步驟“設(shè)、列、解”缺一不可,寫出函數(shù)關(guān)系式及定義域后用導(dǎo)數(shù)求最值,解決最值問題.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)

9、航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四【變式訓(xùn)練1】 一艘輪船在航行時每小時的燃料費和它的速度的立方成正比.已知速度為每小時10海里時,燃料費是每小時6元,而其他與速度無關(guān)的費用是每小時96元,問輪船的速度是多少時,航行1海里所需的費用總和最小?MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典

10、例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYAN

11、LIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四【例2】 某廠生產(chǎn)某種電子元件,如果生產(chǎn)出一件正品,那么可獲利200元,如果生產(chǎn)出一件次品,那么損失100元.已知該廠制造電子元件過程中,次品率p與日產(chǎn)量x的函數(shù)關(guān)系是 (xN+),(1)將該廠的日盈利額T(單位:元)

12、表示為日產(chǎn)量x(單位:件)的函數(shù);(2)為獲最大盈利,該廠的日產(chǎn)量應(yīng)定為多少件?分析:根據(jù)次品率,先計算出正品數(shù)與次品數(shù),再用獲利總數(shù)減去損失總數(shù)可得盈利.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳

13、理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITO

14、UXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四反思反思解決此類有關(guān)利潤的實際應(yīng)用題,應(yīng)靈活運用題設(shè)條件,建立利潤的函數(shù)關(guān)系,常見的基本等量關(guān)系有:(1)利潤=收入-成本;(2)利潤=每件產(chǎn)品的利潤銷售件數(shù).MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知

15、識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLI

16、TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUIT

17、ANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四【例3】 如圖,要設(shè)計一張矩形廣告,該廣告含有大小相等的左右兩個矩形欄目(即圖中陰影部分),這兩欄的面積之和為18 000 cm2,四周空白的寬度為10 cm,兩欄之間的中縫空白的寬度為5 cm.怎樣確定廣告的高與寬的尺寸(單位:cm),能使矩形廣告的面積最小?分析:設(shè)出適當(dāng)?shù)淖兞?把廣告面積用該變量表示出來,然后用導(dǎo)數(shù)解答最值問題.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITAN

18、GYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練Z

19、HISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四令S0,得x140;令S0,得20 x140.函數(shù)在(140,+)上是增加的,在(20,140)上是減少的,S(x)的最小值為 S

20、(140).當(dāng)x=140時,y=175,即當(dāng)x=140,y=175時,S取得最小值24 500 cm2.故當(dāng)廣告的高為140 cm,寬為175 cm時,可使廣告的面積最小.反思反思解決面積、容積的最值問題,要正確引入變量,將面積或容積表示為變量的函數(shù),結(jié)合實際問題的定義域,利用導(dǎo)數(shù)求解函數(shù)的最值.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIA

21、N隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四【變式訓(xùn)練3】 用長為90 cm,寬為48 cm的長方形鐵皮做一個無蓋的容器,先在四個角分別截去一個小正方形,再把四邊翻轉(zhuǎn)90,然后焊接成如圖所示的容器,問該容器的高為多少時,容器的容積最大?最大容積是多少?MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SH

22、ULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四解:設(shè)容器的高為x cm,容器的容積為V(x)cm3.則V(x)=x(90-2x)(48-2x)=4x3-276x2+4 320 x(0 x24)

23、.V(x)=12x2-552x+4 320=12(x2-46x+360)=12(x-10)(x-36)(0 x24).令V(x)=0,得x1=10,x2=36(舍去).當(dāng)0 x0,V(x)是增加的;當(dāng)10 x24時,V(x)0,V(x)是減少的.因此,在定義域(0,24)內(nèi),函數(shù)V(x)只有在x=10處取得最大值,其最大值為V(10)=10(90-20)(48-20)=19 600.故當(dāng)容器的高為10 cm時,容器的容積最大,最大容積是19 600 cm3.MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)

24、導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四易錯點:應(yīng)用函數(shù)模型時,不注意實際問題的定義域而致錯【例4】 現(xiàn)有一批貨物由海上從A地運往B地,已知輪船的最大航行速度為35海里/時,A地至B地之間的航行距離約為5

25、00海里,每小時的運輸成本由燃料費和其余費用組成,輪船每小時的燃料費與輪船速度的平方成正比(比例系數(shù)為0.6),其余費用為每小時960元.(1)把全程運輸成本y(單位:元)表示為速度x(單位:海里/時)的函數(shù).(2)為了使全程運輸成本最小,輪船應(yīng)以多大速度行駛?MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識

26、梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLIT

27、OUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三題型四MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITA

28、NGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHI

29、SHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航

30、DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 53已知某生產(chǎn)廠家的年利潤y(單位:萬元)與年產(chǎn)量x(單位:萬件)之間的函數(shù)關(guān)系式為y=- x3+81x-234,則使該生產(chǎn)廠家獲取最大年利潤的年產(chǎn)量為()A.13萬件B.11萬件C.9萬件D.7萬件解析:y=-x2+81,令y=0,得x=9(x=-9舍去),且經(jīng)討論知x=9是函數(shù)的極大值點,也是最大值點,故廠家獲得最大年利潤的年產(chǎn)量是9萬件.答案:CMUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理

31、目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典

32、例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5MUBIAODAOHANG目標(biāo)導(dǎo)航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5