In the video at the beginning of the lesson, you saw a sample dog run. Breeders of guide dogs for the blind need dog runs to make sure their young dogs stay safe, healthy, and happy as they train. These spaces are typically a fenced in area outside. If a dog run needs to be 3 feet less than 2 times the width of the run, then how much fencing will you need to build one?
To calculate the amount of fencing, you'll need to calculate the perimeter of this rectangular space since you are putting in a fence that travels all the way around the edges of the dog run. The formula is P = 2L + 2W.
First, let's sketch a diagram of the dog run. In your sketch, include labels for the length and width with the dimensions described here. You'll need to write expressions for both the length and the width in order to label the sides correctly. Once drawn, click the Answer button to compare diagrams.
The problem indicates that a "dog run needs to be 3 feet less than 2 times the width of the run," so in the diagram width = w and length = 2w-3.
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To find the perimeter, substitute into the formula, P = 2L + 2W, and solve. L represents length and W represents width. We just wrote the expression 2w-3 for length and w for width, so they get substituted in. |  |
P = 2L + 2W P = 2(2w - 3) + 2w |
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At this point, you cannot determine either the P or the w--you'll need one of them as an actual number in order to solve. But you can simplify the expression 2(2w - 3) using the distributive property. | |
2(2w - 3) = 4w - 6 |
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Substitute that back into P = 2(2w - 3) + 2w. | |
P = (4w - 6) + 2w |
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But are you done simplifying? There are still two like terms, 4w and 2w, that can be combined. Order of operations indicates parenthesis first and then add, but inside the parenthesis you cannot combine the terms, so the parenthesis become meaningless and can be removed. | |
P = 4w - 6 + 2w = 6w - 6 |
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You now have an expression for the perimeter of the fenced in area. You've used the distributive and commutative properties to simplify. | |
6w - 6 |
