Asked by Madeline
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)
Answers
Answered by
Steve
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.
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.
Answered by
nicole
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
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