Lesson #3: Maximizing Area and Minimizing Perimeter

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Presentation transcript:

Lesson #3: Maximizing Area and Minimizing Perimeter Unit 3: Geometry Lesson #3: Maximizing Area and Minimizing Perimeter

LEARNING GOALS To find the maximum area given the perimeter and show work by using sketches, words and numbers. To find the minimum perimeter given the area and show work by using sketches, words and numbers.

A measurement of length in one direction DIMENSIONS A measurement of length in one direction Examples: width, depth and height are dimensions. 80 m 40 m 40 x 120 by 80 x 80 120 m by 80 m

MAX AREA

Think about planning a garden layout, designing a parking lot, fencing off a cornfield, or enclosing a swimming area in a lake. All these situations involve MAXIMIZING the area of a rectangle.

Think about planning a garden layout, designing a parking lot, fencing off a cornfield, or enclosing a swimming area in a lake. All these situations involve MAXIMIZING the area of a rectangle. Imagine you are helping to organize an outdoor concert. You need to use a number of moveable barriers to design an enclosed region for the crowd. For the comfort and safety of the fans, the enclosure must provide as much area as possible. Which barrier set up is the best? Some possible designs are shown: Wide and shallow Square-like Narrow and deep

To find the maximum or BEST area... If you are given a certain perimeter, all you have to do is take the perimeter that they gave you and divide it by 4. For example, if Farmer Sara has 320 m of fencing and wants to use it to make the largest space possible for her pigs to live, she will have to divide 320 by 4. You will get 80, so each side will be 80 m.

Farmer Ben gets a length of fence of 320 m Farmer Ben gets a length of fence of 320 m. He has to make some kind of rectangular yard to enclose his animals. So it has to be 320 m in perimeter. He is stuck to 320 m as his perimeter. He comes up with three different rectangles that he thinks would be good.

80 m A= 4800 m2 40 m Square-like A= 6400 m2 120 m 80 m A= 6000 m2 60 m WHICH ONE HAS THE MOST AREA THAT YOU CAN GET OUT OF A PERIMITER OF 320 m? 100 m When p= 320 m, the max area is 80 x 80 m and it is a square. One of these shapes has most possible area you can get if you enclose them rectangularly!

Note – All Four Sides The maximum area for a rectangular shape for a given perimeter is ALWAYS a square. You want to get the best BANG FOR YOUR BUCK, so if you are given a perimeter and asked to find the maximum area, you will always give a square area. If you are given a perimeter 16 m, divide it by 4 to get the dimensions. 4m x 4 m= 16 m If you are given a perimeter of 28 cm, divide by 4 to get the dimensions… 7 cm by 7 cm = 49 cm2

KEY CONCEPTS Different rectangles with equal perimeters can enclose DIFFERENT amounts of area. It is often useful to determine the dimensions of the rectangle that has maximum area. The dimensions of a rectangle that has maximum area depend on the number of sides to be fences. “Natural fencing,” such as rivers, can increase the amount of enclosed area.

Mat’s Questions: A farmer wants to construct a fenced rectangular yard. Suggest 2 pieces of advice to maximize the enclosed area. A) When does a square-shaped rectangle maximize the enclosed area? B)When does a square-shaped rectangle NOT maximize the enclosed area? You need to fence only three sides of a rectangle. How are the maximum area length and width related in this case?

EXAMPLE If a father has 100 m of fencing, what is the maximum rectangular area he can enclose for his children to play in?

We need to think of all the dimensions of a rectangle that can be made with a Perimeter of 100 m and see which dimensions give the biggest area. P = 2l + 2w and A =lw. The easiest way to do this is to prepare and complete a table that gives length, width and area for a rectangle with a perimeter of 100 m. Because a rectangle’s opposite sides are equal, we know that the l and w must add up to ½ the perimeter – in this case 50 m. If a father has 100 m of fencing, what is the maximum rectangular area he can enclose for his children to play in?

You might also want to sketch the various rectangles to help visualize the results, such as: A= 225 m2 5 m 45 m A= 525 m2 15 m 35 m

Width Length Area = lw 5 45 5(45) = 225 10 40 10(40) = 400 15 35 15(35) = 525 20 30 20(30) = 600 25 25(25) = 625 30(20) = 600 35(15) = 525 40(10) = 400 45(5) = 225

Homework Page 70-71, Q #1-4, 6

MIN PERIMETER

If you are given the area, the best way to figure out what the least amount of perimeter will be needed is to take the area measurement and press the square root button on your calculator.  For example, if you are given an area of 400 m2, you press the square root button on your calculator. You will get 20. So each side will be 20 m. To find the least amount of perimeter that will be needed to have this area, simply add the sides. 20+20+20+20 or multiply 20 by 4. Either way, you will get a perimeter of 80 m.  The most cost-efficient way to maximize area and minimize perimeter will always be a SQUARE!

A= 400 m2 Farmer Ben has been given an area he needs to cover: 400 m2 He doesn’t want to spend a lot of money on fencing. One of these areas has the least amount of fencing required. Length and Width have to multiply to become 400. 10 x 40 = 400 A= 400 m2 10 m P = 40 m + 40 m + 10 m + 10 m= 100m 40 m

1) 10 x 40 = 400 A= 400 m2 10 m P = 40 m + 40 m + 10 m + 10 m= 100m 40 m 2) 25 x 16 = 400 A= 400 m2 16 m P = 25 m + 25 m + 16 m + 16 m= 82 m 25 m 2) 20 x 20 = 400 A= 400 m2 20 m P = 20 m + 20 m + 20 m + 20 m= 80 m 20 m JUST TAKE THE SQUARE ROOT OF THE AREA TO DETERMINE THE DIMENSIONS√400

EXAMPLE A father can get a building permit to enclose 1024 m2 for a pool area. What is the smallest length of fence he would need to put in to keep his costs the lowest?

ONLY 3 SIDES!

CHECK THIS OUT! http://www.youtube.com/watch?v=bddoE0DG4JU

Notes – Only Three Sides The maximum area and minimum perimeter occur when the length (side parallel to the side without the border) is twice the width. l = 2w

Example The City of Ottawa wants to fence off a swimming area at Moonie’s Bay Beach. Find the optimal value given the information below. The maximum swimming area that can be enclosed with 220m of rope. The minimum rope needed to enclose a 3200m2 swimming area.

Example - Max Area Perimeter (m) Width (m) Length (m) Area (m2) 220 5 210 1050 10 200 2000 20 180 3600 30 160 4800 40 140 5600 50 120 6000 60 100 55 110 6050

Example – Min. Perimeter Area (m2) Width (m) Length (m) Perimeter (m) 3200 10 320 340 20 160 200 30 106.7 166.7 40 80 50 64 164 confirm 35 91.4 161.4 45 71.1 161.1

SUCCESS CRITERIA FOR THIS LESSON I can find the perimeter of a rectangle and square. I can identify the formula for the area of a rectangle and square. I can explain what the square root of a number is. I can use my calculator to find the square root of a number. I can explain what finding the minimum perimeter means. I can find the smallest amount of perimeter needed to enclose something (like a fence), if given an area. I can draw a shape that displays the dimensions of the minimum perimeter. I can explain what finding the maximum area means. I can explain the purpose of finding the maximum area. I can find the largest amount of space that can be had if I am given the perimeter of a rectangle. I can draw a diagram that displays the dimensions of the maximum area.