A model used for the yield Y of an agricultural crop as a function of the nitrogen level N in the soil (measured in appropriate units) is

Question

Y = KN/(9+N^2)

where k is a positive constant. What nitrogen level gives the best yield?

N=

Answer 1

This problem asks us to optimize the yield (Y) of a crop as a function of the nitrogen level in the soil. To do this optimization, we must first use the quotient rule to find the derivative of the function given. The quotient rule is as follow:

If h(x) = f(x)/g(x), then h'(x) = (f'(x)*g(x) - f(x)*g'(x))/(g(x))²

Using our given equation, we find the derivative like so:

Y = (kN)/(9 + N²)

Y' = ((k)(9 + N²) - (kN)(2N))/(9 + N²)

Simplifying, we get:

Y' = ((9k + kN²) - (2kN²))/(N⁴ + 18N² +81)

Y' = (9k - kN²)/(N⁴ + 18N² +81)

Y' = k(9 - N²)/(N⁴ + 18N² +81)

The next step to solving this is setting the derivative equal to 0 as it will give us the values of the local maxima and minima of the equation. To do this, we simply set the numerator equal to zero, solve for values of N, check to make sure they do not also make the denominator equal to 0, and then plug the numbers into the original equation to see which gives us the maximum value.

Setting the numerator equal to 0 we get:

k(9 - N²) = 0

9 - N² = 0

9 = N²

N = 3, -3

Plugging the terms into our denominator, we find that both give it a nonzero value of 324, meaning they are valid solutions. You could have also recognized that since none of the terms in the denominator could ever be negative since they're all being raised to even exponents, you didn't even need to check this part since the minimum value of the denominator is 81, or well above 0. Plugging both 3 and -3 into our initial equation, we find 2 answers:

Y(3) = 3k/18

Y(3) = k/6

Y(-3) = -3k/18

Y(-3) = -k/6

Seeing that this answer is in terms of k, we cannot simplify it any further since k is not given. Thus, there are two possible answers. If k is positive, then a nitrogen value of 3 gives the maximum. If k is negative, then a maximum is achieved at a nitrogen value of 3. However, were not done yet. Since nitrogen is a real-life value, and you can never have a negative amount of a substance, you can conclude that the answer must be 3 and not -3. Thus, to achieve a maximum yield of crop, the nitrogen value must be equal to 3 units.