Question

Imagine you are an engineer for a water-treatment plant, and you must determine the optimal amount of water in the plant over a 24-hour period.
On a particular day, the amount of untreated water coming into the plant can be modeled by () = 100 + 30cos⁡,, where t is in hours since midnight and f(t) represents thousands of gallons of water. The amount of treated water at any given time, t, can be modeled by () = 30cos⁡(/2).
a) Define a new function, ′(), that would represent the amount of untreated water inside the plant, at any given time, t.
b) Find ′().
c) Determine the critical values of this function over the interval [0, 24).
d) Determine whether the critical values represent local maximums or minimums.
e) Determine the maximum and minimum amount of untreated water in the plant for the day.

Answer 1

The critical values are t-values where the derivative of the function is not defined or where it is 0 . Then we must solve $a^{\prime}(t)=0$ :

$a^{\prime}(t)=0=-5 \sin \left(\frac{t}{6}\right)+15 e^{\cos (t 2)} \sin \left(\frac{t}{2}\right)$

What is meant by local maximums?

A point (x, y) on the graph of a function is said to be a local maximum point if its y coordinate is larger than all other y coordinates at positions "near to" it (x, y).

The following function can be used to simulate how much untreated water enters a facility:

$f(t)=100+30 \cos \frac{t}{6}$

This same plant's water treatment capacity is shown by:

$g(t)=30 e^{\cos \frac{t}{2}}$

To indicate the amount of untreated water present inside the plant at any given time, t, we must define the function a(t) in section (a). This amount equals the amount of raw water entering the plant less the amount of water being treated concurrently. To sum up, we have:$a(t)=f(t)-g(t)=100+30 \cos \frac{t}{6}-30 e^{\cos \frac{t}{2}}$

We must determine its derivative in section (b). Let's review how to derive an exponential expression and a cosine in order to accomplish that:$h(t)=\cos (k(t)) \Rightarrow h^{\prime}(t)=-\sin (k(t)) \cdot k^{\prime}(t)$

$b(t)=e^{c(t)} \Rightarrow b^{\prime}(t)=e^{c(t)} \cdot c^{\prime}(t)$

Using these properties we can find a'(t):

$a^{\prime}(t)=f^{\prime}(t)-g^{\prime}(t)=\left(100+30 \cos \frac{t}{6}\right)^{\prime}-\left(30 e^{\cos \frac{t}{2}}\right)^{\prime}$

$a^{\prime}(t)=-\frac{30}{6} \sin \frac{t}{6}-30 e^{\cos \frac{t}{2}} \cdot\left(-\frac{1}{2} \sin \frac{t}{2}\right)$

$a^{\prime}(t)=-5 \sin \frac{t}{6}+15 e^{\cos \frac{1}{2}} \sin \frac{t}{2}\end{gathered}$

In part (c) we must find the critical points of the function over [0, 24). The critical values are t-values where the derivative of the function is not defined or where it is 0 . Then we must solve $a^{\prime}(t)=0$ :

$a^{\prime}(t)=0=-5 \sin \left(\frac{t}{6}\right)+15 e^{\cos (t 2)} \sin \left(\frac{t}{2}\right)$

The complete question is:

On a particular day, the amount of untreated water coming into the plant can be modeled by f(t) = 100 + 30cos(t/6) where t is in hours since midnight and f(t) represents thousands of gallons of water. The amount of treated water at any given time, t, can be modeled by g(t) = 30e^cos(/2)a) Define a new function, a(t), that would represent the amount of untreated water inside the plant, at any given time, t.b) Find a′ (t).c) Determine the critical values of this function over the interval [0, 24).d) Determine whether the critical values represent local maximums or minimums.e) Determine the maximum and minimum amount of untreated water in the plant for the day.

To learn more about local maximum and minimum refer to:

https://brainly.com/question/11894628

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