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Whitepunk [10]
3 years ago
14

A landscape architect plans to enclose a 3000 square feet rectangular region in a botanical garden. She will use shrubs costing

25 dollar per foot along three sides and fencing costing 15 dollars per foot along the fourth side. Find the dimensions of the botanical garden that will minimize the total cost. Follow the steps:
Mathematics
1 answer:
Ainat [17]3 years ago
6 0

Answer:

20L + 15L + 15(2W) = Cost = C

LW = 3000

W=3000/L

35L + 30(3,000/L) = C

C(L) = 35L + 90,000L^-1

take the derivative of C(L) and set equal to zero, solve for L

C' = 35 - 90000L^-2 = 0

35=90,000/L^2

L^2 = 90,000/35

L= 300/sqr35 = 300sqr35/35= about 50.71 feet of fencing one 1 side and of shrubs on the opposite side

W =3000/300/sqr35 = 10sqr35=59.16 feet of shrubs on 2 sides

W=59.2

L=50.7

WL = square feet

but rounding to one decimal gives WL=3001 square feet

at cost of 30(59.2) + 15(50.7) + 20(50.7)

= 1776 + 1774.5 = $3550.50

but 30(60)+35(50)=1800+1750= $3550.00, 50 cents less

and 60x50=3000 square feet

30(59.16)+35(50.71) = 1774.8 + 1774.85 = $3549.65 the minimum cost with 59.16 by 50.71 feet

rounding errors make a few cents difference

more exact dimensions are 59.16079783 by 50.70925528 feet for minimum cost

for calculus on the absolute minimum, take the 2nd derivative

or look at the end points

60x50 feet is virtually the cost minimizing dimensions.

look at some other simple numbers on either side and you'll find higher cost.  It's

a local minimum, but the endpoints show it's also an absolute minimum

Step-by-step explanation:

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Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 50, \sigma = 6

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

Z = \frac{X - \mu}{\sigma}

Z = \frac{67 - 50}{6}

Z = 2.83

Z = 2.83 has a pvalue 0.9977

X = 60

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

Z = \frac{X - \mu}{\sigma}

Z = \frac{76 - 50}{6}

Z = 4.33

Z = 4.33 has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

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At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

Z = \frac{X - \mu}{\sigma}

-1.28 = \frac{X - 50}{6}

X - 50 = -1.28*6

X = 42.32

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

Z = \frac{X - \mu}{\sigma}

-1.28 = \frac{X - 50}{6}

X - 50 = -1.28*6

X = 42.32

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{X - 50}{6}

X - 50 = 1.28*6

X = 57.68

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

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