Exercise 60 Page 12

Practice makes perfect

This exercise gives us a lot of freedom, there are literally thousands of rectangles that have a perimeter of 400 feet. Let's explore a series of them. It can be more efficient to get the length and width by thinking about the perimeter of half of the rectangle.

cc P&=&2l+2w &⇕ P/2&= &2l+2w/2 &⇕ 400/2&=&l +w In our case, the perimeter is 400 meaning half of that would be 200. We can choose numbers for our length and width that add to 200. Let's draw some rectangles.

Keep in mind those are just four possible solutions.

We need to find the area of each rectangle from Part A, plus an additional rectangle. Let's put it in the suggested table. Recall that length and width add up to 200.

Rectangle	Length	Width	Area
1	160 ft	40 ft	6400 ft^2
2	150 ft	50 ft	7500 ft^2
3	120 ft	80 ft	9600 ft^2
4	x ft	200-x ft	x(200-x) ft^2

For this part, we need to make a graph that relates the width to the area. Let's add a row to our table from Part B and get some more data points to see if there is pattern.

Length (x)	10	20	30	40	50	60	70	80	90	100	110	120	130	140	150	160	170	180	190	x
Width	190	180	170	160	150	140	130	120	110	100	90	80	70	60	50	40	30	20	10	200-x
Area (y)	1900	3600	5100	6400	7500	8400	9100	9600	9900	10 000	9900	9600	9100	8400	7500	6400	5100	3600	1900	200x-x^2

Now let's look at the graph of the equation y=200x-x^2.

The peak of the graph is the largest possible area, and that is when y = 10 000.

We can use the largest possible area from Part C, 10 000 ft^2 and use the table to determine the length and width that gave us that area.

Length	50	60	70	80	90	100
Width	150	140	130	120	110	100
Area	7500	8400	9100	9600	9900	10 000

This table indicates that both the length and the width are 100 ft when the area is at its maximum.

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