Problem 71 - Single Variable Calculus

Question

A Mixture Problem 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 \(x\) 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 approximately 5.714 lb, which occurs at \(t \approx 6.1744\) minutes.

Step 1: Find the first derivative of the function

First, we want to find the first derivative of the function \(f(t) = 1.5(10-t) - 0.0013(10-t)^4\). To do this, we will apply basic calculus rules: $$ f'(t) = \frac{d}{dt}(1.5(10-t) - 0.0013(10-t)^4) $$ $$ f'(t) = -1.5 - 4 \times 0.0013(10-t)^3 = -1.5 - 0.0052(10-t)^3 $$

Step 2: Find the critical points

We now want to find the critical points of the function where the first derivative is either equal to 0 or undefined. In this case, f'(t) is a polynomial and is defined for all values of t. So, we will find when f'(t) = 0: $$ -1.5 - 0.0052(10-t)^3 = 0 $$ Solve for \(t\): $$ 0.0052(10-t)^3 = 1.5 $$ $$ (10-t)^3 = \frac{1.5}{0.0052} $$ $$ t = 10 - \sqrt[3]{\frac{1.5}{0.0052}} $$ The critical point is approximately \(t \approx 6.1744\).

Step 3: Second derivative test

To determine if the critical point corresponds to a maximum or a minimum, we will use the second derivative test. We find the second derivative of the function: $$ f''(t) = \frac{d^2}{dt^2}(-1.5 - 0.0052(10-t)^3) $$ $$ f''(t) = -0.0156(10-t)^2 $$ Now, we check the sign of the second derivative at the critical point \(t \approx 6.1744\): $$ f''(6.1744) = -0.0156(10-6.1744)^2 < 0 $$ Since the second derivative is negative, the critical point corresponds to a local maximum.

Step 4: Check the endpoints

Next, we need to check the endpoints to make sure the local maximum value is the absolute maximum in the interval \(0 \leq t \leq 10\). At \(t = 0\): $$ f(0) = 1.5(10-0) - 0.0013(10-0)^4 = 15 - 0.0013(10000) = 2 $$ At \(t = 10\): $$ f(10) = 1.5(10-10) - 0.0013(10-10)^4 = 0 $$ The local maximum value is higher than both of the endpoint values of the function. Therefore, the maximum amount of salt present in the tank at any time is at \(t \approx 6.1744\) minutes. We can find the actual maximum salt amount by substituting the critical point back into the function: $$ f(6.1744) \approx 1.5(10-6.1744) - 0.0013(10-6.1744)^4 \approx 5.714 \mathrm{lb} $$ Thus, the maximum amount of salt present in the tank at any time is approximately 5.714 lb.

Key Concepts

Differentiation

In calculus, differentiation is a fundamental concept that deals with finding the rate at which a function is changing at any given point. It involves computing the derivative of a function, which is essentially a measure of its instantaneous rate of change.
To differentiate a function, you apply rules such as the power rule, product rule, and chain rule. These rules help simplify the process of finding derivatives for various types of functions. Differentiation is particularly useful in finding slopes and rates, which are crucial for solving real-world problems.

In our exercise, we first find the derivative of the function given in the problem. The function represents the amount of salt in the tank over time. By calculating its first derivative, we can understand how the salt quantity changes as time progresses.

Mixture Problems

Mixture problems in mathematics often deal with combining two or more substances to form a mixture. They require understanding rates of flow and concentration of each component in the mixture. This type of problem is common in contexts like chemistry, physics, and environmental science.
Typically, mixture problems involve equations that express the relationship between the concentration and volume of substances entering and leaving a container over time. Calculating these changes helps in determining how the mixture's composition evolves.

In our exercise, the tank starts with a specific concentration of salt which changes as brine enters and exits the tank. The rate at which brine flows in and out means that the concentration of salt at any time can vary, making it essential to find the maximum concentration during the process.

Critical Points

Critical points in calculus are where the first derivative of a function is zero or undefined. These points are crucial because they help identify possible maxima, minima, or points of inflection in the function.
To find critical points, solve the equation where the first derivative equals zero. Once found, these points can be further examined to understand their nature, specifically whether they are points of maximum or minimum value for the function.

For our exercise, we identified a critical point using the derivative we found. This critical point is crucial because it helps determine at what time the quantity of salt in the tank is at its highest.

Second Derivative Test

The second derivative test is a method to classify critical points found using the first derivative. It involves evaluating the second derivative of the function at these critical points.
If the second derivative at a critical point is positive, the point is a local minimum. If negative, the point is a local maximum. This test helps affirm whether the critical point is indeed a peak or a valley on the graph of the function.

In our problem, the second derivative was calculated to confirm that the critical point is a local maximum. This confirms the point in time when the salt concentration reaches its peak in the tank.