## 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.

## 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 $) Connections (millions) 2006 38 59.1 2007 43.9 65.7 2008 49.3 72.1 2009 60.3 96.5 2010 63.4 102.2 2011 70.2 107.8 Let’s examine the relationship between connections and revenue by graphing them in a scatter plot. If the independent variable is the number of connection in millions and the dependent variable is the corresponding revenue in billions or dollars, we get the scatter plot below. A cubic model for this data is$latex \displaystyle R(c)=0.00047{{c}^{3}}-0.11885{{c}^{2}}+10.31365c-254.48905$When we place the model on the scatter plot, we see that the model fits the data well. The cubic model has an inflection point. On the left side of the inflection point, the revenue is rising at a slower and slower rate. This means the slopes of tangent lines get smaller as they move from left to right near the inflection point. The graph is concave down on the left side of the inflection point. On the right side of the inflection point, the graph increases faster and faster. In other words, the graph gets steeper and steeper. This results in the graph being concave up on the right side of the inflection point. To locate the inflection point, we need to track the concavity of the function using a second derivative number line. Concavity may change anywhere the second derivative is zero. The first and second derivatives are$latex \displaystyle {R}'(c)=0.00141{{c}^{2}}-0.23770c+10.31365latex \displaystyle {R}”(c)=0.00282c-0.23770$Set the second derivative equal to zero and solve for x:$latex \displaystyle 0.00282c-0.23770=0latex \displaystyle c=\frac{0.23770}{0.00282}\approx 84.3$Label this point on a second derivative number line. Now test points on either side, at 50 and 100, to find the sign of the second derivative.$latex \displaystyle {R}”(50)=0.00282(50)-0.23770<0latex \displaystyle {R}”(100)=0.00282(100)-0.23770>0\$

Label the number line with – and + to indicate the sign of the second derivative. Now we can label the concavity on the number line. 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. 