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

55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.