2. Crop yield A model used for the yield $$ Y $$ of an agricultural crops as a function of the nitrogen level $$ N $$ in the soil is $$ Y=\frac{27 N}{1+N^{2}} $$ What nitrogen level gives the best yield?

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We are given $$ Y=\frac{27N}{1+N^2}. $$ **Step 1. Compute the derivative $$ Y' $$ using the quotient rule.** Let $$ u(N)=27N $$ and $$ v(N)=1+N^2 $$. The quotient rule states: $$ Y'=\frac{u'(N)v(N)-u(N)v'(N)}{[v(N)]^2}. $$ Calculate: $$ u'(N)=27, $$ $$ v'(N)=2N. $$ Thus, $$ Y'=\frac{27(1+N^2)-27N(2N)}{(1+N^2)^2}. $$ **Step 2. Simplify the derivative.** Simplify the numerator: $$ 27(1+N^2)-54N^2 = 27-27N^2. $$ So, $$ Y'=\frac{27(1-N^2)}{(1+N^2)^2}. $$ **Step 3. Find the critical points by setting $$ Y'(N)=0 $$.** We set the numerator equal to zero: $$ 27(1-N^2)=0. $$ Divide both sides by 27 (non-zero): $$ 1-N^2=0. $$ Thus, $$ N^2=1 \quad \Longrightarrow \quad N=1 \quad \text{or} \quad N=-1. $$ Since $$ N $$ represents the nitrogen level in the soil, it must be nonnegative. Therefore, we dismiss $$ N=-1 $$. **Step 4. Verify that $$ N=1 $$ gives a maximum yield.** Examine the sign of $$ Y' $$ around $$ N=1 $$. - For $$ N<1 $$: $$ 1-N^2>0 $$, so $$ Y'(N)>0 $$ (increasing). - For $$ N>1 $$: $$ 1-N^2<0 $$, so $$ Y'(N)<0 $$ (decreasing). Since $$ Y $$ increases before $$ N=1 $$ and decreases after $$ N=1 $$, $$ N=1 $$ corresponds to a maximum. **Step 5. Compute the best yield.** Substitute $$ N=1 $$ back into the model: $$ Y=\frac{27 \cdot 1}{1+1^2}=\frac{27}{2}. $$ Thus, the best yield is $$ \frac{27}{2} $$, which is attained when $$ N=1 $$. The nitrogen level that gives the best yield is $$ N = 1 $$, resulting in a yield of $$ \frac{27}{2} $$.

2. Crop yield A model used for the yield \( Y \) of an agricultural crops as a function of the nitrogen level \( N \) in the soil is \[ Y=\frac{27 N}{1+N^{2}} \] What nitrogen level gives the best yield?

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