Larosa machine shop (lms) is studying where to locate its tool bin facility on the shop floor. the locations of the five production stations appear in figure shown below location y demand fabrication 1.0 4.0 12 paint 1.0 2.0 19 subassembly 1 subassembly 2 assembly 2.5 2.0 13 3.05.0 22 4.0 4.0 in an attempt to be fair to the workers in each of the production stations, management has decided to try to find the position of the tool bin that would minimize the sum of the distances from the tool bin to the five production stations. we define the following decision variables: x horizontal location of the tool bin y vertical location of the tool bin we may measure the straight line distance from a station to the tool bin located at (xm by using euclidean (straight-line) distance. for example, the distance from fabrication located at the coordinates 4) to the tool bin located at the coordinates (kc7 is given by ax-dr-4) (a) suppose we know the average number of daily trips made to the tool bin from each productiorn station. the average number of trips per day are 12 for fabrication, 19 for paint, 13 for subassembly 1, 7 for subassembly 2 and 22 for assembly. it seems like we would want the tool bin closer to those stations with high average numbers of trips. develop a new unconstrained model that minimizes the sum of the demand-weighted distance defined as the product of the demand (measured in number of trips) and the distance to the station. min: (b) solve the model you developed in part (a). if required, round your answer to six decimal places. do not round intermediate calculation. (c) the solution to the un-weighted model is x= 2.230 and γ= 3.349. comment on the differences between the unweighted distance solution given and the demand-weighted solution found in part the input in the box below will not be graded, but may be reviewed and considered by your instructor.

p> q if you calculate p is100 q is 79

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

yes the data was not prsented

step-by-step explanation: