Mr.jones bought 20 yards of fencing to make pen for dog pen will be in rectangle shape, the fencing comes in 1 yard section, wants the MAXIMUM areafor his dog to run, what dimension will give the pen the biggest space for dog to run?? then what r the dimensions for the maximum area if he buys 32 yards of fencing? Make table of widths, lengths, and areas. what would be the demensions of the maximum area if he buys 28 yards of fencing and make table showing widths, lengths and areas????
Sorry I am normally great at Pre-Algebra, but this one i cannot figure out. Sorry, Kim! Good Luck!
To find the dimensions that will give the maximum area for the dog pen, we can use the properties of rectangles.
Let's start with the first scenario where Mr. Jones has 20 yards of fencing.
1. Set up an equation: The perimeter of a rectangle is given by P = 2(length + width). Since Mr. Jones has 20 yards of fencing, this means 20 = 2(length + width).
2. Solve for one variable: We can rearrange the equation to get width in terms of length: width = (20 - 2length)/2 = 10 - length.
3. Substitute the value: Substitute this expression for width into the formula for the area of a rectangle: Area = length * width = length * (10 - length).
4. Determine the maximum area: To find the dimensions that maximize the area, we need to find the length for which the area is maximum. We can do this by graphing the equation or by finding the vertex of the quadratic function. In this case, since the coefficient of the quadratic term is negative (-1), we know that the vertex will give us the maximum area. The x-coordinate of the vertex can be found using the formula: x = -b/2a, where a = -1 and b = 10. Plugging in these values, we get x = -10/(-2) = 5.
So, when Mr. Jones has 20 yards of fencing, the dimensions that give the maximum area for the dog pen are length = 5 yards and width = 10 - length = 10 - 5 = 5 yards. The maximum area is therefore Area = length * width = 5 * 5 = 25 square yards.
Now, let's move on to the second scenario where Mr. Jones has 32 yards of fencing.
1. Set up the equation: Similar to the previous scenario, we have 32 = 2(length + width).
2. Solve for width: Again, rearrange the equation to get width in terms of length: width = (32 - 2length)/2 = 16 - length.
3. Substitute the value: Substitute this expression for width into the formula for the area: Area = length * width = length * (16 - length).
4. Determine the maximum area: Using the same process as before, we find the x-coordinate of the vertex for this quadratic function. Using the formula x = -b/2a, where a = -1 and b = 16, we get x = -16/(-2) = 8.
So, when Mr. Jones has 32 yards of fencing, the dimensions that give the maximum area for the dog pen are length = 8 yards and width = 16 - length = 16 - 8 = 8 yards. The maximum area is therefore Area = length * width = 8 * 8 = 64 square yards.
Now, let's consider the last scenario where Mr. Jones has 28 yards of fencing.
1. Set up the equation: Once again, we start with 28 = 2(length + width).
2. Solve for width: Rearrange the equation to get width in terms of length: width = (28 - 2length)/2 = 14 - length.
3. Substitute the value: Substitute this expression for width into the formula for the area: Area = length * width = length * (14 - length).
4. Determine the maximum area: Using the formula x = -b/2a, where a = -1 and b = 14, we find x = -14/(-2) = 7.
Therefore, when Mr. Jones has 28 yards of fencing, the dimensions that give the maximum area for the dog pen are length = 7 yards and width = 14 - length = 14 - 7 = 7 yards. The maximum area is therefore Area = length * width = 7 * 7 = 49 square yards.
Here is a table summarizing the dimensions and areas for the three scenarios:
| Fencing (yards) | Length (yards) | Width (yards) | Area (square yards) |
|-----------------|----------------|---------------|---------------------|
| 20 | 5 | 5 | 25 |
| 32 | 8 | 8 | 64 |
| 28 | 7 | 7 | 49 |
I hope this explanation helps you understand how to find the dimensions that give the maximum area for the dog pen based on the amount of fencing available.