The number of salmon swimming upstream to spawn is approximated by the following function:
S(x) = -x^3 + 2x^2 + 405x + 4965 where X represents the temperature of the water in degrees Celsius and (6 <= x <= 20
Find the critical value(s) of S(x).
Using the critical values that fall within the domain, apply the 2nd derivative test for the max/min to find the water temperature that produces the maximum number of salmon swimming upstream. (Show work for the 2nd derivative).
The water temperature that produces the maximum number of salmon swimming upstream is ____degrees Celsius.

Let [tex]S(x) = -x^{3}+2\cdot x^{2}+405\cdot x +4965[/tex], for [tex]6 \leq x \leq 20[/tex]. [tex]x[/tex] represents the temperature of the water, measured in degrees Celsius, and [tex]S[/tex] is the number of salmon swimming upstream to spawn, dimensionless.

We compute the first and second derivatives of the function:

[tex]S'(x) = -3\cdot x^{2}+4\cdot x +405[/tex] (Eq. 1)

[tex]S''(x) = -6\cdot x +4[/tex] (Eq. 2)

Then we equalize (Eq. 1) to zero and solve for [tex]x[/tex]:

[tex]-3\cdot x^{2}+4\cdot x +405 = 0[/tex]

And all roots are found by Quadratic Formula:

[tex]x_{1} \approx 12.305\,^{\circ}C[/tex], [tex]x_{2}\approx -10.971\,^{\circ}C[/tex]

Only the first root is inside the given interval of the function. Hence, the correct answer is:

[tex]x \approx 12.305\,^{\circ}C[/tex]

Now we evaluate the second derivative at given result. That is:

[tex]S''(12.305) = -6\cdot (12.305)+4[/tex]

[tex]S''(12.305) = -69.83[/tex]

According to the Second Derivative Test, a negative value means that critical value leads to a maximum. In consequence, the water temperature that produces the maximum number of salmon swimming upstream is approximately 12.305 degrees Celsius.