A farmer needs to build a rectangular corral for his animals. He has 200 yards of fencing available. He needs to make 4 pens. What is the largest corral he can create? (Remember the pens also count as a part of the perimeter, not just the outside of the corral)
depends on the configuration of the pens. A longs skinny corral with 4 pens in a row? Or a rectangular corral with a 2x2 arrangement of pens?
If the first, then we have two lengths and 5 widths.
2x+5y = 200
area = xy = x(200-2x)/5 = -2/5 x^2 + 40x
the vertex of the parabola is at x=40/(4/5) = 50
So, the corral is 50 by 25
If the pens are in a grid, then we have 2 length and 3 widths:
2x+3y = 200
area = xy = x(200-2x)/3 = -2/3 x^2 + 200/3 x
The vertex is now at x=50, so the corral is 50 by 33.33
Looks like the square-ish pen is bigger.
write the equation for the line that passes through the point (-4,7) and is perpendicular to 3y-8=x-4. leave equation in thw point-slope form show all work
To find the largest corral the farmer can build, we need to determine the dimensions of each pen and the overall shape of the corral.
Since there are 4 pens, we can divide the available fencing equally to create 4 sides for each pen. Let's assume the length of each pen is 'l' and the width is 'w'.
We know that the perimeter of each pen will be its length multiplied by 2 plus its width multiplied by 2:
Perimeter of each pen = 2l + 2w
Since there are 4 pens in total, the total perimeter of all the pens together would be:
Total Perimeter = 4(2l + 2w)
Now, we also need to accommodate the overall shape of the corral. The corral will have the same length as each pen and its width will be the width of one pen plus the width of the adjacent pen:
Overall Length = l
Overall Width = 2w
The total perimeter of the corral will include the perimeter of the pens as well as the outside perimeter of the corral:
Total Perimeter = Perimeter of all the pens + Perimeter of the corral
Given that the total available fencing is 200 yards, we have the equation:
200 = 4(2l + 2w) + 2l + 2w
Simplifying the equation further:
200 = 8l + 8w + 2l + 2w
200 = 10l + 10w
To maximize the size of the corral, we can assume that the length and width should be as equal as possible. Let's assume that l = w.
Substituting l = w into the equation:
200 = 10l + 10l
200 = 20l
l = 10
Now, we know that the length of each pen is 10 yards. Substituting this value into the equation for the width:
200 = 10(2) + 10w
200 = 20 + 10w
10w = 200 - 20
10w = 180
w = 18
Therefore, the largest corral the farmer can create is a rectangular corral with overall dimensions of 10 yards by 18 yards, consisting of 4 pens each measuring 10 yards in length and 2 pens measuring 18 yards in width.