step 1\nwe need to maximize ( y(n)=\frac{k n}{36+n^{2}} ) for ( n geq 0 ).\n( y^{prime}(n)=\frac{left(36+n^{2…

step 1\nwe need to maximize ( y(n)=\frac{k n}{36+n^{2}} ) for ( n geq 0 ).\n( y^{prime}(n)=\frac{left(36+n^{2}\right)(k)-left(k n\right)(2 n)}{left(36+n^{2}\right)^{2}} )\nstep 2\nafter simplifying, we have ( y^{prime}(n)=\frac{kleft(36-n^{2}\right)}{left(36+n^{2}\right)^{2}} ). since ( y^{prime} ) is never undefined, we need only find where ( y^{prime}=0 ), which occurs where the numerator is 0. the only ( n ) value for ( n geq 0 ) which makes the numerator 0 is ( n= )

step 1\nwe need to maximize ( y(n)=\frac{k n}{36+n^{2}} ) for ( n geq 0 ).\n( y^{prime}(n)=\frac{left(36+n^{2}\right)(k)-left(k n\right)(2 n)}{left(36+n^{2}\right)^{2}} )\nstep 2\nafter simplifying, we have ( y^{prime}(n)=\frac{kleft(36-n^{2}\right)}{left(36+n^{2}\right)^{2}} ). since ( y^{prime} ) is never undefined, we need only find where ( y^{prime}=0 ), which occurs where the numerator is 0. the only ( n ) value for ( n geq 0 ) which makes the numerator 0 is ( n= )

Answer

Explanation:

Step1: Set the numerator equal to zero

We have (k(36 - N^{2}) = 0). Since (k\neq0) (otherwise (Y(N)=0) for all (N) and there is no non - trivial maximum), we solve (36 - N^{2}=0).

Step2: Solve the equation for (N)

[ \begin{align*} 36 - N^{2}&=0\ N^{2}&=36\ N&=\pm6 \end{align*} ] Since (N\geq0), we take (N = 6).

Answer:

(6)