急!!!求解一道AP微积分应用题!
Themanagerofalargeapartmentcomplexknowsfromexperiencethat120unitswillbeoccupiedifther...
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. 展开
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. 展开
1个回答
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dN/dx = -1/3
dN = -(1/3)dx
N = -(1/基州3)x + C
Initial condition: N=120=-(1/3)350 + C
C = 710/山念3
N = -x/3 + 710/搏唯蔽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/基州3)x + C
Initial condition: N=120=-(1/3)350 + C
C = 710/山念3
N = -x/3 + 710/搏唯蔽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
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