## What Kind Of Tools Can Help Me To Calculate Riemann Sums?

In Sections 13.2 and 13.3, you will be calculating areas using an approximate methods called Riemann Sums. For small numbers of data points or small numbers of rectangles, we can easily calculate a Riemann Sum by hand. However, as the number of rectangles gets larger (like more than 8 rectangles) the task becomes overwhelming. Luckily, there are online calculators that make the task trivial.

To be able to use this calculator, you need to know the formula for the function f (x), where the sums will run, the number of rectangles, and whether the rectangle will touch the function on the left or right hand side.

In the image above, the function we are finding the Riemann sum for is f (x) = 2x+1 and we are forming rectangles from x = 1 to x = 4. In this case we have chosen to use 3 rectangles that touch on the right side of the rectangles. This type of Riemann Sum would be referred to as a Right Hand Sum (RHS).

If we were to have the rectangles touch on the left hand side, we would have a Left Hand Sum (LHS). In this case we would change the “taking the samples at the Right” to “taking the samples at the Left”

Make sure you choose Replot after you make any changes.

We can double the number of rectangles to 6 to get

If you continue to increase the number of rectangles with LHS or RHS, the estimate of the area will get closer and closer to the actual area (which we can find using geometry).

Use this tool in your homework to help relieve the drudgery of adding up all of the sums. Keep in mind that if you are given data points or a graph, you will have to work out the sums by hand.

## 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 $4 times the maximum stores in the warehouse at any time, we get 59520 ∙ 4. Now let’s fill in the second row: I hope you see the pattern emerging. Let’s fill in the last row: The total cost function in ? is the sum of all of the individual costs or After simplifying this function, you can find its minimum by taking the derivative and setting equal it equal to zero. This will give you the critical number that you can then verify using the second derivative. ## How Do You Minimize The Surface Area Of A Container? As the maximization and minimization problems become more complicated, using a table to organize examples of what is being modeled becomes more and more useful. Let’s look at another example of how this might work. Problem A rectangular container with a square base, an open top, and a volume of 8788 ft3 is to be constructed of sheet metal. Find the dimensions of the tank that has the minimum surface area. ## How Is Area Related To A Definite Integral? You might have seen the marketing campaign for the Super Bass-o-Matic 76. Let’s look at some options for finding the increased costs when changing production of Super Bass-o-Matic blenders from 200 blenders to 800 blenders. Suppose the marginal cost is given by$latex \displaystyle {C}'(x)=30-0.02x {\text{ dollars per blender}}$By finding the area of rectangles, we get values that indicate an increase in cost. For instance, the first term of the left hand sum below,$latex \displaystyle \left( 26\frac{\$}{\text{blender}} \right)\left( 100\text{ blenders} \right)=\$2600$This means that if we estimate the cost per blender as 26 dollars per blender for all 100 blenders from 200 to 300 blenders, the increased cost is$2600. If we continue this for more rectangles until 800, the sum of the areas give an estimate for the increased cost from increasing production from 200 to 800.

In this case the left hand sum is $11400 and the right hand sum is$12600…the difference between these estimates is $1200. If we do the the same process for 60 rectangles (each being 10 wide), we get additional estimates. In this case, the left hand sum is$11940 and the right hand sum is 12060. The difference between the estimate for 60 rectangles is $120. Ten times as many rectangles leads to a tenth of the difference. We get a similar picture for 600 rectangles. The left hand sum is$11994 and the right hand sum is $12006. In this case the difference between the estimates is$12. Also notice that as the number of rectangles gets larger, the area in the rectangles looks more and more like the area under the function and above the horizontal axis.

Let’s summarize what happens as the number of rectangles increases.

 Number of Rectangles LHS RHS Difference 6 11400 12600 1200 60 11940 12060 120 600 11994 12006 12

You might guess that 6000 rectangles would lead to a difference of $1.2 and you would be correct. The left hand sum is$11999.4 and the right hand sum is $12000.6. In fact, as the number of rectangles gets larger and larger (each rectangle getting narrower and narrower) each estimate gets closer and closer to$12000.

This is the exact area in the blue shaded region under the function, above the horizontal axis, and between 200 and 800. To convince yourself that this is the case, we can this area by breaking the region up into a rectangle and triangle.

The area is

$latex \displaystyle 14\cdot 600+\frac{1}{2}\cdot 12\cdot 60=12000$

The exact area under the function is represented with a definite integral,

$latex \displaystyle \int\limits_{200}^{800}{(30-0.02x)\,dx}=12000$

## How Do You Find Special Points on a Parabola?

Let’s look at how to use formulas for a parabola to get certain important points on a parabola.

Problem For the parabola y = 2x2 + 3x – 2, locate the points below.

a. The y-intercept.

Solution At the y-intercept, the x value is zero. This means that we need to set x = 0 in the equation:

y = 2(02) + 3(0) – 2 = -2

Putting this together, the y-intercept is at (0, -2).

b. The vertex.

Solution The vertex is located using the formula   where the values of a, b, and c come from the equation. In this case, a = 2, b = 3, and c = -2. This gives an x value on the intercept of

To find the corresponding y value, put this value into the equation,

This means the vertex is at (-3/4, –25/8).

c. The x-intercepts.

Solution At the x-intercepts, the y value is zero. Putting this into the equation yields

0 = 2x2 + 3x – 2

This equation is solved with the quadratic formula,

Put the values from the equation (a = 2, b = 3, and c = -2),

The x intercepts are at (-2, 0) and (1/2, 0).

All of these points are shown in the graph of the parabola below.