University:
High School
Course:
High School Mathematics
Academic year:

2023
Views:

11

Pages:

4
Author:

anon

Optimal Value of Measurements (Two Dimensions) Optimization is the process of finding values that make a given quantity the greatest (least) possible given certain conditions. Maximum - greatest possible Minimum – least possible Example 1 (Maximize Area Given Perimeter): a) Suppose you have 40 m of fencing. You want to make a rectangular dog pen. What is the maximum area you can provide for a dog? Rectangl e 1 2 3 4 5 6 7 8 9 10 11 12 13 Width(m) Length(m) Area(𝑚2) Perimeter(m) 1 2 3 4 5 6 7 8 9 10 11 12 13 19 18 17 16 15 14 13 12 11 10 9 8 7 19 36 51 54 75 84 91 96 99 100 99 96 91 40 40 40 40 40 40 40 40 40 40 40 40 40 ∴ The maximum area we can provide for the dog is 100𝑚 2 b) What dimensions give us the maximum area? The dimensions that give us the maximum area are 10m by 10m So, the shape that gives us the maximum area is a square. Example 2: Workers at a resort set up a rectangular area to store outdoor equipment and furniture. They use metal stands. They have 26 stands, each 3 m long. The storage enclosure they set up could have different shapes. How many stands should be used for the width and length to make the largest possibleenclosure? Number of stands along the width Number of stands along the length Width of enclosure (m) Length of enclosure (m) Area enclosed ( 1 2 3 4 5 6 7 8 12 11 10 9 8 7 6 5 3 6 9 12 15 18 21 24 36 33 30 27 24 21 18 15 108 148 270 324 360 378 378 360 2 𝑚) ∴ There should be used & stands along the and 6 stands along the width or vice versa. OR 𝑃𝑠𝑞𝑢𝑎𝑟𝑒 = 45 24 4 = 6 2 Example 3: Suppose we have a rectangle with an area of 24 𝑢𝑛𝑖𝑡𝑠 . What dimensions will give us the minimum perimeter? What shape will minimize the perimeter Rectangle Width(m) Length(m) Area 2 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 8 12 24 24 12 8 6 4.8 4 3 2 1 (𝑚 ) 24 24 24 24 24 24 24 24 24 Perimeter (m) 50 28 22 20 19.6 20 22 28 50 The shape that provides the minimum perimeter is almost a square. Conclusion: For a rectangular enclosure with a given perimeter, if fencing is required on all sides, the optimal are occurs when the enclosure is a square Example 4: You have 80 metres of fencing. What dimensions will give the largest possible area? What is the maximum area? Solution: Given: P=80m Aim for a square. 𝑃 = 45 ← 𝑠𝑢𝑏. 𝑃 = 80 80 = 45 ← 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑠 80 4 = 45 4 20=s ∴ s=20 ∴ The dimensions that will give us the largest possible area are 20m by 20m 2 𝐴𝑠𝑞𝑢𝑎𝑟𝑒 = 𝑆 = (20) 2 = (20)(20) = 400 ∴ The maximum area is 400𝑚 2