How Do You Find the Optimal Size Production Run?

Batch size problems are easily solved if the total cost function is given to you. Since this is a minimization problem at its heart, taking the derivative to find the critical point and then applying the first of second derivative test does the trick. However, if we need to find the total cost function the problem is more involved. In Section 12.4 I pointed out a strategy in which we can apply our pattern recognition skills and a table to come up with the total cost function. Let’s look at an example:

Problem Every year, Danielle Santos sells 59,520 cases of her Delicious Cookie Mix.  It costs her $2 per year in electricity to store a case, plus she must pay annual warehouse fees of $4 per case for the maximum number of cases she will store.  If it costs her $747 to set up a production run, plus $6 per case to manufacture a single case, how many production runs should she have each year to minimize her total costs?

Solution Start the table by labeling the columns:

I have filled in the table with some examples of number of production runs and size of production run to establish the pattern in the last row. Now let’s fill in the first row. Examine each of the entries in the first row carefully.

You can click on the table to view a larger image.

This is mostly the same as in the text. However, the warehouse fees are new. Since the fees are $4 times the maximum stores in the warehouse at any time, we get 59520 ∙ 4.

Now let’s fill in the second row:

I hope you see the pattern emerging. Let’s fill in the last row:

The total cost function in ? is the sum of all of the individual costs or

After simplifying this function, you can find its minimum by taking the derivative and setting equal it equal to zero. This will give you the critical number that you can then verify using the second derivative.

How Do You Minimize The Surface Area Of A Container?

As the maximization and minimization problems become more complicated, using a table to organize examples of what is being modeled becomes more and more useful. Let’s look at another example of how this might work.

Problem A rectangular container with a square base, an open top, and a volume of 8788 ft3 is to be constructed of sheet metal. Find the dimensions of the tank that has the minimum surface area.

Continue reading “How Do You Minimize The Surface Area Of A Container?”

How Do You Find The Equation Of A Tangent Line?

A tangent line to a function is a line that looks most like the function at a point. In common terms, it just grazes the function.

To find its equation, we need to locate the point where the two meet as well as the slope of the function at that point. Then we can use the slope-intercept form or point-slope form of a line to get the equation.

Find the equation of the tangent line to

at x = 3.

Since this problem is asking for the equation of a line, let’s start with the point-slope form

This requires a point (x1, y1) and slope m. We’ll use the function to get the point and the derivative to get the slope of the tangent line.

Find the point: We are given a point x = 3. To find the corresponding y value, put the x value into the function

Find the slope of the tangent line: We need h′(3) to get the slope of the tangent line. We’ll use the Power Rule to take the derivative,

The slope of the tangent is

Write the equation of the tangent line: Putting the point (3, 10) and the slope 9 into the line yields

If you are asked to write this in slope-intercept form, you’ll need to solve this for y to give

If you graph h(x) and the tangent line together, it should be obvious that your tangent line is correct (ie. tangent).

How Do You Calculate A Rate From A Function?

In many business problems, we are interested in knowing the rate at which some quantity is changing. If this quantity is modeled by a function, the rate is modeled by the derivative of the function.

In the problem below, we are given a model of revenue for Verizon, a national wireless carrier. To find the rate, we’ll need to take the derivative of the revenue function.

From 2006 through 2011, Verizon Wireless grew steadily. As the number of connections increased, so did the revenue from those connections.A cubic model for the revenue over this period is

where c is the number of connections in millions.

At what rate is the revenue increasing when there are 80,000,000 connections?

Questions that ask about rate or refer to “marginal” are questions about the derivative. In this case, it is referring to the value of the derivative at c = 80. The derivative of the revenue function is

The value of the derivative at 80 million connections is

To find the units on this rate, recall that the derivative is also the slope of the tangent line at c = 80. The slope of this line has vertical units of billions of dollars and horizontal units of millions of connections. This may be simplified to

Since a billion divided by a million is one thousand. So means that an additional connection at this level results in an increase in revenue of 0.31265 thouusand dollars or $321.65.

What Are Points Of Inflection and How Do You Find Them?

From 2006 through 2011, Verizon Wireless has grown steadily. The table below illustrates the growth in revenue as well as connections (phone numbers).

Year

Revenue (billions $)

Connections (millions)

2006

38

59.1

2007

43.9

65.7

2008

49.3

72.1

2009

60.3

96.5

2010

63.4

102.2

2011

70.2

107.8

Let’s examine the relationship between connections and revenue by graphing them in a scatter plot. If the independent variable is the number of connection in millions and the dependent variable is the corresponding revenue in billions or dollars, we get the scatter plot below.

verizon_scatter

A cubic model for this data is

\displaystyle R(c)=0.00047{{c}^{3}}-0.11885{{c}^{2}}+10.31365c-254.48905

When we place the model on the scatter plot, we see that the model fits the data well.

verizon_model

The cubic model has an inflection point. On the left side of the inflection point, the revenue is rising at a slower and slower rate. This means the slopes of tangent lines get smaller as they move from left to right near the inflection point. The graph is concave down on the left side of the inflection point.

On the right side of the inflection point, the graph increases faster and faster. In other words, the graph gets steeper and steeper. This results in the graph being concave up on the right side of the inflection point.

 To locate the inflection point, we need to track the concavity of the function using a second derivative number line. Concavity may change anywhere the second derivative is zero. The first and second derivatives are

\displaystyle {R}'(c)=0.00141{{c}^{2}}-0.23770c+10.31365

\displaystyle {R}''(c)=0.00282c-0.23770

Set the second derivative equal to zero and solve for x:

\displaystyle 0.00282c-0.23770=0

\displaystyle c=\frac{0.23770}{0.00282}\approx 84.3

Label this point on a second derivative number line.

number_line_6

Now test points on either side, at 50 and 100, to find the sign of the second derivative.

\displaystyle {R}''(50)=0.00282(50)-0.23770<0

\displaystyle {R}''(100)=0.00282(100)-0.23770>0

Label the number line with – and + to indicate the sign of the second derivative.

number_line_5

Now we can label the concavity on the number line.

number_line_7

Since the concavity changes at approximately 84.3, it corresponds to an inflection point. Its position on the graph is found by substituting 84.3 into the original function.

revenue_eval_1