A farmer uses x lb of fertilizer per acre at a cost of 1.4 per pound, leading to a revenue of R = 700 – 400e -x/100 dollars per acre. (a) How many pounds of fertilizer should be applied per acre to maximize profit? Write you answer in decimal form rounding to three decimal places
Number
pounds per acre.
(b) What is the maximum profit on a 200 acre farm?
$ Number

The cost can be written as:

C(x) = $1.4*x

The revenue can be written as:

R(x) = $700 - $400*e^(-x/100)

And as you know, the profit is written as the difference between the revenue and the cost:

P(x) = R(x) - C(x)

P(x) =  $700 - $400*e^(-x/100) - $1.4*x

Now we want to maximize this, then we must look at the zeros in the derivate of P(x)

dP/dx = P'(x) = (-1/100)*$400*e^(-x/100) - $1.4

The maximum will be when P'(x) = 0.

we can solve:

0 = (-1/100)*$400*e^(-x/100) - $1.4

$1.4 = (-1/100)*$400*e^(-x/100)

(-100)*$1.4 = $400*e^(-x/100)

$140 = $400*e^(-x/100)

($140/$400) = e^(-x/100)

0.35 = e^(-x/100)

Now we can apply the Ln() in both sides and get:

ln(0.35) = ln(e^(-x/100) ) = -x/100

-ln(0.35)*100 = x = 104.98

You should apply 104.90 pounds of fertilizer.

b) The profit per acre, is:

P(x) = $700 - $400*e^(-x/100) - $1.4*x

Then if you have 200 acres, the profit will be:

200*P(x) = 200*($700 - $400*e^(-x/100) - $1.4*x)

and the maximum profit is whit x = 104.98 pounds, then:

200*P(104.98) = 200*($700 - $400*e^(-104.98/100) - $1.4*104.98) = $82,604.98

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation:

to determine the unit cost your would divide the total cost by the number sold.

7 bagels for $4.41       $4.41/7 = $.63 each at fresh side

5 bagels for $3.60         $3.60/5 = $.72 each at value mart