# 2題微積分的問題

1.the total cost of producing and selling 100x units of a particular commodity per week is

C(n)=1000+33x-9x^2+x^3

find (a) the level of production at which the marginal cost is a minimum, and (b) the

minimum marginal cost.

2.for th price function given by p(x)= (800/(x+3))-3

find the number of units x1 that makes the total revenue a maximum and state the

maximum possible revenue. What is the marginal revenue when the optimum number of

units,x1, is sold?

Rating

1.Total cost of producing and selling 100 units of a particular commodity per week is C(n)=1000+33x-9x^2+x^3. Find (a) level of production at which the marginal cost is a minimumAns:Cost: C(x)=100(x^3-9x^2+33x+1000)\$/unit.weekMarginal cost: C'(x)=300(x^2-6x+11)Find min: 0=C"=2x-6 => x=3 (units).....ans

(b) the minimum marginal cost.Ans:min C'=C'(3)=300(9-18+11)=600(\$/week.unit^2).....ans

2.Price function given by P(x)=[800/(x+3)]-3. Find (a) the number of units x=? making a maximal total revenue Ans:R(x)=revenue=Price*Quantity=[800/(x+3)-3]x=800x/(x+3)-3x=> 800x=(R+3x)(x+3)=xR+3R+3x^2+9xFind max for x=? letting R'=0:{800x}'={xR+3R+3x^2+9x}'800=xR'+R+3R'+6x+9.....(1)=R+6x+9=800x/(x+3)-3x+6x+9791=800x/(x+3)+3x791(x+3)=800x+3x(x+3)0=3x^2+(809-791)x-3*791=3x^2+18x-3*791=x^2+6x-791x=-3+√800=-3+20√2=25.3=> 25 (uits)

(b) state the maximum possible revenue. Ans:Max=R(25)=[800/(x+3)-3]x=(800/28-3)25=(200/7-3)25=(200-21)25/7=179*25/7=639.3(\$).....ans

(c) What is the marginal revenue when the optimum number of x units is sold?Ans:Marginal revenue = Eq.(1): R'(x)=(791-6x-R)/(x+3)=(791-6x)/(x+3)-[800/(x+3)-3]x/(x+3)=(791-6x)/(x+3)-[800x/(x+3)^2]+3x/(x+3)=(791-3x)/(x+3)-800x/(x+3)^2R'(25)=[(791-25)28-800*25]/28^2=(766*28-20000)/28^2=1448/28*28=181/7*14=1.847(\$/unit)

• Anonymous
6 years ago