搜搜考试网求解一道AP微积分应用题!
来源：学生作业帮 编辑：搜搜考试网作业帮 分类：英语作业 时间：2024/07/06 23:19:06
求解一道AP微积分应用题!
The manager of a large apartment complex knows from experience that 120 units will be occupied if the rent is 350 dollars per month.A market survey suggests that,on average,one additional unit will remain vacant for each 3 dollar increase in rent.Similarly,one additional unit will be occupied for each 3 dollar decrease in rent.
Let the rent on the apartment be x dollars per month,and let N be the number of apartments rented each month,and let R be the revenue (the gross income) brought in each month by the apartment manager.
Write N as a function of x only.N(x)= apartments.
Write Ras a function of x only.R(x)= dollars.
What rent should the manager charge to maximize revenue?
ANSWER:dollars per month.
The manager of a large apartment complex knows from experience that 120 units will be occupied if the rent is 350 dollars per month.A market survey suggests that,on average,one additional unit will remain vacant for each 3 dollar increase in rent.Similarly,one additional unit will be occupied for each 3 dollar decrease in rent.
Let the rent on the apartment be x dollars per month,and let N be the number of apartments rented each month,and let R be the revenue (the gross income) brought in each month by the apartment manager.
Write N as a function of x only.N(x)= apartments.
Write Ras a function of x only.R(x)= dollars.
What rent should the manager charge to maximize revenue?
ANSWER:dollars per month.
dN/dx = -1/3
dN = -(1/3)dx
N = -(1/3)x + C
Initial condition: N=120=-(1/3)350 + C
C = 710/3
N = -x/3 + 710/3
R = Nx = -(x/3)(x-710) = -(x^2)/3 + 710x/3
dR/dx = -2x/3 +710/3
Set dR/dx = 0, x = 355
ANSWER: 355 dollars per month to maximize revenue
dN = -(1/3)dx
N = -(1/3)x + C
Initial condition: N=120=-(1/3)350 + C
C = 710/3
N = -x/3 + 710/3
R = Nx = -(x/3)(x-710) = -(x^2)/3 + 710x/3
dR/dx = -2x/3 +710/3
Set dR/dx = 0, x = 355
ANSWER: 355 dollars per month to maximize revenue