Math Problem Statement

The graph shows a​ city's daily rate of use of electricity​ (in millions of​ kilowatts). Estimate the total daily usage of electricity by summing the area of rectangles. Use the left​ endpoints, then the right​ endpoints, then give the average of those results as the answer. Let the width of each rectangle be 2 hours. Midnight 4 A.M. 8 A.M. Noon 4 P.M. 8 P.M. Midnight

4 8 12 16 20 24 28

Millions of kilowatt-hours

A coordinate system has a horizontal axis labeled with hours from Midnight to Midnight, in increments of 4 hours and a vertical axis labeled Millions of kilowatt-hours from 0 to 24 in increments of 4. A smooth curve rises from right to left, passing through the points with the following y-coordinates, with the hours listed first: Midnight, 11.29; 4 A.M., 12.62; 8 A.M., 16.62; Noon, 23.29, to a maximum at 4 P.M, 29.15, and falls through the following points: 8 P.M., 20.49; Midnight, 13.48. All coordinates are approximate. Question content area bottom Part 1 Begin by finding the estimate of the total energy used on the day shown in the​ graph, by summing rectangles using left endpoints. The width of each interval is 2 hours.

Solution

To estimate the total daily electricity usage using the left endpoints, we can break the 24-hour period into 2-hour intervals and sum the areas of rectangles formed by these intervals.

Given Data

The points you provided for the electricity usage are:

  • Midnight (0:00) - 11.29 million kWh
  • 4 A.M. (4:00) - 12.62 million kWh
  • 8 A.M. (8:00) - 16.62 million kWh
  • Noon (12:00) - 23.29 million kWh
  • 4 P.M. (16:00) - 29.15 million kWh
  • 8 P.M. (20:00) - 20.49 million kWh
  • Midnight (24:00) - 13.48 million kWh

Left Endpoint Calculation

Using the left endpoints, we take the height of the function at the beginning of each interval:

  1. Midnight to 4 A.M. (0-4):

    • Height = 11.29 million kWh
    • Area = 11.29×2=22.5811.29 \times 2 = 22.58 million kWh
  2. 4 A.M. to 8 A.M. (4-8):

    • Height = 12.62 million kWh
    • Area = 12.62×2=25.2412.62 \times 2 = 25.24 million kWh
  3. 8 A.M. to Noon (8-12):

    • Height = 16.62 million kWh
    • Area = 16.62×2=33.2416.62 \times 2 = 33.24 million kWh
  4. Noon to 4 P.M. (12-16):

    • Height = 23.29 million kWh
    • Area = 23.29×2=46.5823.29 \times 2 = 46.58 million kWh
  5. 4 P.M. to 8 P.M. (16-20):

    • Height = 29.15 million kWh
    • Area = 29.15×2=58.3029.15 \times 2 = 58.30 million kWh
  6. 8 P.M. to Midnight (20-24):

    • Height = 20.49 million kWh
    • Area = 20.49×2=40.9820.49 \times 2 = 40.98 million kWh

Total Area (Left Endpoint)

Now we can sum these areas:

Total Area=22.58+25.24+33.24+46.58+58.30+40.98=226.62 million kWh\text{Total Area} = 22.58 + 25.24 + 33.24 + 46.58 + 58.30 + 40.98 = 226.62 \text{ million kWh}

Right Endpoint Calculation

Now, we calculate using the right endpoints. The heights will be taken from the end of each interval:

  1. Midnight to 4 A.M. (0-4):

    • Height = 12.62 million kWh
    • Area = 12.62×2=25.2412.62 \times 2 = 25.24 million kWh
  2. 4 A.M. to 8 A.M. (4-8):

    • Height = 16.62 million kWh
    • Area = 16.62×2=33.2416.62 \times 2 = 33.24 million kWh
  3. 8 A.M. to Noon (8-12):

    • Height = 23.29 million kWh
    • Area = 23.29×2=46.5823.29 \times 2 = 46.58 million kWh
  4. Noon to 4 P.M. (12-16):

    • Height = 29.15 million kWh
    • Area = 29.15×2=58.3029.15 \times 2 = 58.30 million kWh
  5. 4 P.M. to 8 P.M. (16-20):

    • Height = 20.49 million kWh
    • Area = 20.49×2=40.9820.49 \times 2 = 40.98 million kWh
  6. 8 P.M. to Midnight (20-24):

    • Height = 13.48 million kWh
    • Area = 13.48×2=26.9613.48 \times 2 = 26.96 million kWh

Total Area (Right Endpoint)

Now sum these areas:

Total Area=25.24+33.24+46.58+58.30+40.98+26.96=231.60 million kWh\text{Total Area} = 25.24 + 33.24 + 46.58 + 58.30 + 40.98 + 26.96 = 231.60 \text{ million kWh}

Average of the Results

Finally, we find the average of the left and right endpoint estimates:

Average=226.62+231.602=458.222=229.11 million kWh\text{Average} = \frac{226.62 + 231.60}{2} = \frac{458.22}{2} = 229.11 \text{ million kWh}

Summary

  • Left Endpoint Estimate: 226.62 million kWh
  • Right Endpoint Estimate: 231.60 million kWh
  • Average Estimate: 229.11 million kWh

If you have any questions or need further clarification, feel free to ask! Here are some related questions you might consider:

  1. How would the total usage change if we used different width intervals?
  2. Can we analyze the data using more sophisticated numerical methods?
  3. How does electricity usage vary during different times of the day?
  4. What are the implications of this usage for energy policy?
  5. How can this data be visualized for better understanding?

Tip: For more accurate estimates, consider using more intervals or applying numerical integration techniques like trapezoidal or Simpson's rule.

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Math Problem Analysis

Mathematical Concepts

Calculus
Numerical Integration
Riemann Sums

Formulas

Area of Rectangle = height × width
Average = (Left Estimate + Right Estimate) / 2

Theorems

Mean Value Theorem for Integration

Suitable Grade Level

Grades 11-12