中央電大經(jīng)濟(jì)數(shù)學(xué)基礎(chǔ)實(shí)用復(fù)習(xí)題匯總小抄

1、專業(yè)好文檔一、應(yīng)用題1設(shè)生產(chǎn)某種產(chǎn)品個單位時的成本函數(shù)為：（萬元）,求：（1）當(dāng)時的總成本、平均成本和邊際成本；（2）當(dāng)產(chǎn)量為多少時，平均成本最?。拷猓海?）總成本，平均成本，邊際成本 所以，（萬元）， （萬元）（萬元） （2）令 ，得（舍去） 因?yàn)槭瞧湓诙x域內(nèi)唯一駐點(diǎn)，且該問題確實(shí)存在最小值，所以當(dāng)時，平均成本最小. 2.某廠生產(chǎn)某種產(chǎn)品件時的總成本函數(shù)為（元），單位銷售價格為（元/件），問產(chǎn)量為多少時可使利潤達(dá)到最大？最大利潤是多少解：成本為：收益為：利潤為：，令得，是惟一駐點(diǎn)，利潤存在最大值，所以當(dāng)產(chǎn)量為250個單位時可使利潤達(dá)到最大，且最大利潤為（元）。3投產(chǎn)某產(chǎn)品的固定成本為36(

2、萬元)，且邊際成本為(萬元/百臺)試求產(chǎn)量由4百臺增至6百臺時總成本的增量，及產(chǎn)量為多少時，可使平均成本達(dá)到最低解：成本函數(shù)為：當(dāng)產(chǎn)量由4百臺增至6百臺時，總成本的增量為100（萬元），令得，（負(fù)值舍去）。是惟一駐點(diǎn)，平均成本有最小值，所以當(dāng)（百臺）時可使平均成本達(dá)到最低.4已知某產(chǎn)品的邊際成本=2（元/件），固定成本為0，邊際收益，求：產(chǎn)量為多少時利潤最大？在最大利潤產(chǎn)量的基礎(chǔ)上再生產(chǎn)50件，利潤將會發(fā)生什么變化？解：邊際利潤為：令得，。是惟一駐點(diǎn)，最大利潤存在，所以當(dāng)產(chǎn)量為500件時，利潤最大。 - 25（元）即利潤將減少25元。5已知某產(chǎn)品的邊際成本為(萬元/百臺)，為產(chǎn)量(百臺)，固定

3、成本為18(萬元)，求最低平均成本. 解：因?yàn)榭偝杀竞瘮?shù)為 = 當(dāng)= 0時，c(0) = 18，得 c =18，即 c()= 又平均成本函數(shù)為 令 ， 解得= 3 (百臺) 該問題確實(shí)存在使平均成本最低的產(chǎn)量. 所以當(dāng)x = 3時，平均成本最低. 最底平均成本為 (萬元/百臺) 6、已知生產(chǎn)某產(chǎn)品的邊際成本為 (萬元/百臺)，收入函數(shù)為（萬元），求使利潤達(dá)到最大時的產(chǎn)量，如果在最大利潤的產(chǎn)量的基礎(chǔ)上再增加生產(chǎn)臺，利潤將會發(fā)生怎樣的變化？解：邊際利潤為：令得，是惟一駐點(diǎn)，而最大利潤存在，所以當(dāng)產(chǎn)量為3百臺時，利潤最大。當(dāng)產(chǎn)量由3百臺增加到5百臺時，利潤改變量為（萬元） 即利潤將減少4萬元。7.設(shè)

4、生產(chǎn)某產(chǎn)品的總成本函數(shù)為 (萬元)，其中為產(chǎn)量，單位：百噸銷售百噸時的邊際收入為（萬元/百噸），求：利潤最大時的產(chǎn)量；在利潤最大時的產(chǎn)量的基礎(chǔ)上再生產(chǎn)百噸，利潤會發(fā)生什么變化？.解：因?yàn)檫呺H成本為 ，邊際利潤令，得可以驗(yàn)證為利潤函數(shù)的最大值點(diǎn). 因此，當(dāng)產(chǎn)量為百噸時利潤最大. 當(dāng)產(chǎn)量由百噸增加至百噸時，利潤改變量為 （萬元）即利潤將減少1萬元. 8.設(shè)生產(chǎn)某種產(chǎn)品個單位時的成本函數(shù)為：（萬元）,求：當(dāng)時的總成本和平均成本；當(dāng)產(chǎn)量為多少時，平均成本最??？.解：因?yàn)榭偝杀?、平均成本和邊際成本分別為：， 所以， ， 令 ，得（舍去），可以驗(yàn)證是的最小值點(diǎn)，所以當(dāng)時，平均成本最小 。 二、線性代數(shù)計(jì)算

5、題1、 設(shè)矩陣，求。解：因?yàn)?所以，。2、設(shè)矩陣a =，i是3階單位矩陣，求。解：因?yàn)椋?i-a i ) = 所以=。3設(shè)矩陣 a =，b =，計(jì)算(ab)-1解：因?yàn)閍b = (ab i ) = 所以 (ab)-1= 4、設(shè)矩陣，求解：求逆矩陣的過程見復(fù)習(xí)指導(dǎo)p77的4，此處從略 。；所以，。5設(shè)矩陣，求解矩陣方程。解： 6.設(shè)矩陣，求.解：利用初等行變換得 即 由矩陣乘法得。7求線性方程組的一般解解：因?yàn)樵鰪V矩陣 所以一般解為 （其中是自由未知量）8求線性方程組的一般解解：因?yàn)橄禂?shù)矩陣 所以一般解為 （其中，是自由未知量） 9、當(dāng)取何值時，齊次線性方程組有非0解？并求一般解。解：因?yàn)橄禂?shù)矩

6、陣 所以當(dāng)= 4時，該線性方程組有無窮多解，且一般解為： （其中是自由未知量）。10、問當(dāng)取何值時，線性方程組有解，在有解的情況下求方程組的一般解。解：方程組的增廣矩陣所以當(dāng)時，方程組有解；一般解為：（其中是自由未知量）11解：所以，方程組的一般解為：（其中是自由未知量）12求線性方程組.解：將方程組的增廣矩陣化為階梯形 此時齊次方程組化為 得方程組的一般解為其中是自由未知量 13.當(dāng)為何值時，線性方程組有解，并求一般解。解：所以，當(dāng)時，有解。一般為：（其中是自由未知量三、微分計(jì)算題1設(shè)，求解：因?yàn)?所以 2計(jì)算積分 解： 3設(shè)，求解： 4計(jì)算積分 解： 5.設(shè)，求. 解：由導(dǎo)數(shù)運(yùn)算法則和復(fù)合

7、函數(shù)求導(dǎo)法則得 6.計(jì)算 10分 解：由不定積分的湊微分法得 7.已知，求. 解：由導(dǎo)數(shù)運(yùn)算法則和復(fù)合函數(shù)求導(dǎo)法則得 8計(jì)算. 解：由定積分的分部積分法得四、練習(xí)題（1），求解：（2），求解：（3），求解：（4），求解：（5），求解：（6）解：（7）解：（8）解：（9）解：方法1（10）解：（11）解：（12）解：（13）解：（14）解：五、復(fù)習(xí)指導(dǎo)1、設(shè)，求。解：2、設(shè)，求。解：3、設(shè)，求。解：4、設(shè)，求。解：5、設(shè)，求。解：6、設(shè)，求。解：7、設(shè)，求。解：8、解：原式=9、解：原式=10、解：原式=11、解：原式=12、解：原式=if we dont do that it will go

8、on and go on. we have to stop it; we need the courage to do it.his comments came hours after fifa vice-president jeffrey webb - also in london for the fas celebrations - said he wanted to meet ivory coast international toure to discuss his complaint.cska general director roman babaev says the matter

9、 has been exaggerated by the ivorian and the british media.blatter, 77, said: it has been decided by the fifa congress that it is a nonsense for racism to be dealt with with fines. you can always find money from somebody to pay them.it is a nonsense to have matches played without spectators because

10、it is against the spirit of football and against the visiting team. it is all nonsense.we can do something better to fight racism and discrimination.this is one of the villains we have today in our game. but it is only with harsh sanctions that racism and discrimination can be washed out of football

11、.the (lack of) air up there watch mcayman islands-based webb, the head of fifas anti-racism taskforce, is in london for the football associations 150th anniversary celebrations and will attend citys premier league match at chelsea on sunday.i am going to be at the match tomorrow and i have asked to

12、meet yaya toure, he told bbc sport.for me its about how he felt and i would like to speak to him first to find out what his experience was.uefa hasopened disciplinary proceedings against cskafor the racist behaviour of their fans duringcitys 2-1 win.michel platini, president of european footballs go

13、verning body, has also ordered an immediate investigation into the referees actions.cska said they were surprised and disappointed by toures complaint. in a statement the russian side added: we found no racist insults from fans of cska.baumgartner the disappointing news: mission aborted.the superson

14、ic descent could happen as early as sunda.the weather plays an important role in this mission. starting at the ground, conditions have to be very calm - winds less than 2 mph, with no precipitation or humidity and limited cloud cover. the balloon, with capsule attached, will move through the lower l

15、evel of the atmosphere (the troposphere) where our day-to-day weather lives. it will climb higher than the tip of mount everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude of commercial airliners (5.6 miles/9.17 kilometers) and into the stratosphere. as he crosses th

16、e boundary layer (called the tropopause),e can expect a lot of turbulence.the balloon will slowly drift to the edge of space at 120,000 feet ( then, i would assume, he will slowly step out onto something resembling an olympic diving platform.below, the earth becomes the concrete bottom of a swimming

17、 pool that he wants to land on, but not too hard. still, hell be traveling fast, so despite the distance, it will not be like diving into the deep end of a pool. it will be like he is diving into the shallow end.skydiver preps for the big jumpwhen he jumps, he is expected to reach the speed of sound

18、 - 690 mph (1,110 kph) - in less than 40 seconds. like hitting the top of the water, he will begin to slow as he approaches the more dense air closer to earth. but this will not be enough to stop him completely.if he goes too fast or spins out of control, he has a stabilization parachute that can be deployed to slow him down. his team hopes its not needed. instead, he plans to deploy his 270-square-foot (25-square-meter) main chute at an altitude of around 5,000 feet (1,524 meters).in order to deploy this chute successfully, he will have to slow to 172 mph (277 kph). he will have a reserve pa