Problem 78 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A tank initially contains 10 gal of brine with \(2 \mathrm{lb}\) of salt. Brine with \(1.5 \mathrm{lb}\) of salt per gallon enters the tank at the rate of \(3 \mathrm{gal} / \mathrm{min}\), and the well-stirred mixture leaves the tank at the rate of \(4 \mathrm{gal} / \mathrm{min}\). It can be shown that the amount of salt in the tank after \(t\) min is \(x \mathrm{lb}\) where \(x=f(t)=1.5(10-t)-0.0013(10-t)^{4} \quad(0 \leq t \leq 10)\) What is the maximum amount of salt present in the tank at any time?

Step-by-Step Solution

Answer

The maximum amount of salt present in the tank at any time is \(15 \mathrm{lb}\), which occurs at \(t=0\).

Step 1: Find the derivative of f(t)

First, we find the derivative of \(f(t)=1.5(10-t)-0.0013(10-t)^{4}\) with respect to \(t\). This will help us to find the critical points where the function might have a maximum value. Derivative of \(f(t)\) is: \(f'(t)=-1.5-0.0052(10-t)^{3}\)

Step 2: Find the critical points

To find the critical points, we set the derivative equal to zero and solve for \(t\). \(0=-1.5-0.0052(10-t)^{3}\) Rearrange the equation: \(1.5=0.0052(10-t)^{3}\) Now, solve for \(t\): \((10-t)^{3}=\frac{1.5}{0.0052}\) Take the cube root of both sides: \(10-t=\sqrt[3]{\frac{1.5}{0.0052}}\) Finally, find the value of t: \(t=10-\sqrt[3]{\frac{1.5}{0.0052}}\)

Step 3: Evaluate the endpoints and critical points

We need to evaluate the function at the endpoints \(t=0\) and \(t=10\), and at the critical points found in step 2. 1. For \(t=0\): \(f(0)=1.5(10-0)-0.0013(10-0)^{4}=15\) 2. For \(t=10\): \(f(10)=1.5(10-10)-0.0013(10-10)^{4}=0\) 3. For the critical point \(t=10-\sqrt[3]{\frac{1.5}{0.0052}}\): \(f\left(10-\sqrt[3]{\frac{1.5}{0.0052}}\right)=1.5\left(10-\left(10-\sqrt[3]{\frac{1.5}{0.0052}}\right)\right)-0.0013\left(10-\left(10-\sqrt[3]{\frac{1.5}{0.0052}}\right)\right)^{4}\) Now, plug in the value of \(t\): \(f\left(10-\sqrt[3]{\frac{1.5}{0.0052}}\right)=1.5\left(\sqrt[3]{\frac{1.5}{0.0052}}\right)-0.0013\left(\sqrt[3]{\frac{1.5}{0.0052}}\right)^{4}\)

Step 4: Determine the maximum amount of salt

Based on our evaluations in step 3, we can compare and determine the maximum amount of salt present in the tank at any time: 1. At \(t=0\), there are \(15 \mathrm{lb}\) of salt. 2. At \(t=10\), there are \(0 \mathrm{lb}\) of salt. 3. At the critical point \(t=10-\sqrt[3]{\frac{1.5}{0.0052}}\), there are \(1.5\left(\sqrt[3]{\frac{1.5}{0.0052}}\right)-0.0013\left(\sqrt[3]{\frac{1.5}{0.0052}}\right)^{4} \mathrm{lb}\) of salt. Comparing the values at the endpoints and the critical point, we can see that the maximum amount of salt in the tank is \(15 \mathrm{lb}\), which occurs at \(t=0\).

Key Concepts

Differential Calculus

Differential calculus is a branch of mathematics that deals with how things change. It's all about understanding the rate at which one quantity changes with respect to another. In simpler terms, it helps us figure out how fast something is moving or growing at a particular moment.
Derivatives are the fundamental tool in differential calculus. They measure how a function's output changes as its input changes. For example, in our exercise, the function representing the amount of salt in the tank at any given time is denoted by \( f(t) \).
To find the derivative \( f'(t) \), we analyze how the function's value changes as \( t \), the time, changes. A derivative can tell us a lot about a function's behavior, like where it increases, decreases, or where its highest or lowest points might be, also known as the critical points.

Critical Points

Critical points are specific values where a function's slope is zero or undefined. These points can be potential locations for the function's maximum or minimum values.
In our exercise, after finding the derivative \( f'(t) \), the next step was to set it equal to zero to find its critical points. This is because critical points occur where the slope of the tangent line to the curve is horizontal, indicating a possible peak or trough.
We solve \( 0 = -1.5 - 0.0052(10-t)^3 \) to find the critical point for our function. By determining the value of \( t \) that makes the derivative zero, we spotted the critical point where the maximum amount of salt might reside.

Optimization

Optimization is the process of making something as effective or functional as possible. In calculus, it usually involves finding the maximum or minimum values of a function, setting certain parameters or constraints.
For the salt example, we're using optimization to find the maximum amount of salt that the tank can hold at any given time. After identifying the critical points, which are potential candidates, we evaluate them along with the boundary values of the interval (\( t = 0 \) and \( t = 10 \)).
Through comparison, we found that the function achieves its peak at \( t = 0 \), indicating 15 lbs of salt—highlighting the importance of optimization for determining where a system performs best.

Mathematical Modeling

Mathematical modeling involves creating representations of real-world phenomena to predict outcomes or understand scenarios better. It's about translating a real problem into mathematical language to solve it.
In this exercise, the problem of salt concentration over time in a tank is modeled by the function \( f(t) = 1.5(10-t) - 0.0013(10-t)^4 \). This function serves as a model predicting how salt levels change over ten minutes, considering the rates of flow in and out of the tank.
Mathematical models like this allow us to apply calculus to optimize processes or forecast future conditions. By taking complex real-world situations and translating them into mathematical terms, we gain valuable insights and tools to help solve practical problems, evident in determining the maximum salt content in our exercise.