Answer:
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
