Expert: Paul Klarreich - 3/7/2009

Question
Paul,

Can you help me to solve this? I believe this is a math variation problem.


Water Cost

In a specific region of the country, the amount of a customer’s water bill, W, is directly proportional to the average daily temperature for the month, T, the lawn area A and the square root of F, where F is the family size, and inversely proportional to the number of inches of rain, R.

In one month the average daily temperature is 78 degrees and the number of inches of rain is 5.6. If the average family of four who has a thousand square feet of lawn pays $72.00 for water that month, estimate the water bill in the same month for the average family of six who has 1,500 square feet of lawn.


Here is what I have come up with for a solution, am I correct? Any help would be appreciated!

My Solution:

W=KTAsqrtF/R

72=(K)(78)(1000)(sqrt4)/5.6

72=k(78000)(2)/5.6

72=K15600/5.6

72/.0025846154 = K27857.14286/27857.14286

K = .0025846154

Checking work:

W=(.0025846154)(78)(1000)(sqrt4)/5.6

W=72.00000043 rounded to $72.00

Solving W for a family of 6 with 1500 sq. ft. of lawn:

W=(.0025846154)(78)(1500)(sqrt6)/5.6

W=132.2724461

My Final Answer would be:

W=132.2724461 rounded to $132.27

Checking work:

132.2724461 = K(78)(1500)(sqrt6)/5.6

132.2724461/51176.83927 = K51176.83927/51176.83927

K = .0025846154


Thanks in advance.

Jenny  

Answer
Questioner:   Jenny
Category:  Advanced Math
Private:  Yes
 
Subject:  Math Variation Problem
Question:  Paul,

Can you help me to solve this? I believe this is a math variation problem.


Water Cost

In a specific region of the country, the amount of a customer’s water bill, W, is directly proportional to the average daily temperature for the month, T, the lawn area A and the square root of F, where F is the family size, and inversely proportional to the number of inches of rain, R.

In one month the average daily temperature is 78 degrees and the number of inches of rain is 5.6. If the average family of four who has a thousand square feet of lawn pays $72.00 for water that month, estimate the water bill in the same month for the average family of six who has 1,500 square feet of lawn.
......................................................
If:
W = the Water bill
T = the average Temp.[IN WHAT SCALE?]
A = lawn Area
F = Family size (in humans)
k is some unknown constant, which we shall have to find.

then

W = k T A sqrt(F) / R

Subst:  

T = 78  << I have a serious issue with this.
R = 5.6
A = 1000
F = 4
W = 72


72 = k (78) (1000) sqrt(4) / (5.6)

k = 72(5.6)/[78(1000)(2 = sqrt(4))

k = 72*5.6/78000

k = 0.0025846153846153846153846153846154 or so.

Now use F = 6, A = 1500.

W = k(78)(1500)(sqrt(6))/5.6

W = 132.27244611029161730265334003412, about.

LOOKS GOOD.

P.S. don't mark 'Private' on questions.  I change it anyway.
-----------------------


Here is what I have come up with for a solution, am I correct? Any help would be appreciated!

My Solution:

W=KTAsqrtF/R

72=(K)(78)(1000)(sqrt4)/5.6

72=k(78000)(2)/5.6

72=K15600/5.6

72/.0025846154 = K27857.14286/27857.14286

K = .0025846154

Checking work:

W=(.0025846154)(78)(1000)(sqrt4)/5.6

W=72.00000043 rounded to $72.00

Solving W for a family of 6 with 1500 sq. ft. of lawn:

W=(.0025846154)(78)(1500)(sqrt6)/5.6

W=132.2724461

My Final Answer would be:

W=132.2724461 rounded to $132.27

Checking work:

132.2724461 = K(78)(1500)(sqrt6)/5.6

132.2724461/51176.83927 = K51176.83927/51176.83927

K = .0025846154


Thanks in advance.

Jenny