A model for the yield, Y of N
for an agricultural crop is a function
of the nitrogen level N in the soil.
This model is Y of N equals kN divided by
the sum of 49 and N squared, where k is a positive constant.
Which nitrogen level maximizes or gives the best yield?
To maximize Y of N, we need to determine
their critical numbers,
which is where the first derivative
is equal to zero or undefined.
To determine Y prime of N,
we will have to apply the Quotient Rule.
For the derivative of Y of N,
the denominator is equal to the square
of the denominator of Y of N,
which gives us the square of 49 plus N squared.
And the numerator is a denominator
of the quantity 49 plus N squared
times the derivative of the numerator
which is the derivative of kN with respect to N.
Remember here, we're treating k as a constant.
Then we have minus the numerator of kN
times the derivative of the denominator,
which is a derivative of 49 plus N squared
with respect to N.
And now let's find the derivatives.
We have Y prime of N equals
we'll leave the denominator the same,
which is a square of 49 plus N squared.
In the numerator we have the quantity 49 plus N squared
times derivative of kN with respect to N
which is k times 1 or just k.
Then we have -kN
times the derivative of 49 plus N squared with respect to N,
which is 2N.
And now, let's simplify the numerator.
Distributing k we have 49k
and then plus kN squared,
and then minus 2k N squared.
Combining like terms,
we have 49k minus k N squared.
This is still over the square of 49 plus N squared.
And now we need to determine where the derivative
is equal to zero or undefined.
There are no values of N where we have division by zero
or the derivative is undefined.
And therefore, we find the critical numbers
by setting the derivative equal to zero and solving.
Irrational expression is equal to zero
when the numerator is zero and the denominator is non-zero,
which means we determine the critical numbers
by setting the numerator equal to zero and solving for N.
So we have 49k
minus k N squared equals zero.
Let's factor out the common factor of k,
which gives us k times the quantity 49 minus N squared
which is a difference of squares.
Factoring, we have k times the quantity 7 plus N
times the quantity 7 minus N.
7 plus N is equal to zero,
when N is equal to -7.
And 7 minus N is equal to zero
when N is equal to positive 7.
But remember, N is the nitrogen level
which can't be negative.
So we can exclude N equals -7.
We only have one critical number, which is N equals 7.
From here we need to determine whether Y of N changes
from increasing to decreasing at N equals 7,
meaning the first derivative changes from
positive to negative at N equals 7.
And let's do this using the table above.
Again, assuming N can't be negative.
The first interval for N will be from 0 to 7.
And the second interval will be from 7 to infinity.
Now we'll pick a test value in each interval.
Let's select N equals 6 and N equals 8.
And next step is to determine the sign of Y prime of N
when N is 6 and N is 8.
Y prime of 6 is equal to 49k
minus k times N squared,
where N squared is 6 squared to 36
giving us -36k,
divided by the quantity
the square of 49 plus N squared,
which gives us the square of 49 plus 36,
which is 13k divided by the square of 85.
We can stop here.
Since k is positive, this quotient is positive
or greater than zero.
The first derivative is positive over this interval
and therefore Y of N is increasing over this interval.
And I will determine the sign of Y prime of 8
which is equal to 49k minus k times N squared
is now k times 8 squared or 64k
divided by the square of 49 plus N squared is 64.
Simplifying, we have -15k divided by the square of 113.
Again, we can stop here.
We know k is positive.
So because we have a negative here
this quotient is negative or less than zero.
And therefore, the derivative is negative
over this interval indicating Y of N is decreasing
over this interval.
So now we know the function changes
from increasing to decreasing at N equals 7,
and therefore, we give a relative maximum.
So N equals 7 is a nitrogen level
that will maximize or give the best yield.
Before we go, let's verify our result by graphing Y of N.
Here I'm using Desmos to graph, Y of N,
but I did replace N with x
so we have Y equals kx
divided by the quantity 49 plus x squared.
And that created a slider here for k.
If you look at the graph on the right,
notice how we do have a maximum at x equals 7
which is really N equals 7.
And as we change the value of k,
notice how the location or the nitrogen level
does not change that maximizes the yield.
So this graph does verify N equals 7,
where a nitrogen level of 7
will maximize the crop yield.
I hope you found this helpful.