SOLUTION: You have two square plots formed with 100m of fencing. Determine the equation of the curve of best fit, and use the equation to determine the value of x that minimizes the area of

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: You have two square plots formed with 100m of fencing. Determine the equation of the curve of best fit, and use the equation to determine the value of x that minimizes the area of       Log On


   

Question 975755: You have two square plots formed with 100m of fencing. Determine the equation of the curve of best fit, and use the equation to determine the value of x that minimizes the area of each square.
(Must use quadratics, no derivatives/calculus)
Only thing I've seen to do is A+=+X%5E2+%2B+Y%5E2 then 100+=+4x+%2B+4y. 25=x%2By, y=25-x.
A=X%5E2+%2B+%2825-x%29%5E2
A=x%5E2+%2B+625+-+25x+-+25x+%2B+x%5E2
A=x%5E2+-+50x+%2B+625
But then solutions for x are imaginary, what do I do now?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
You have two square plots formed with 100m of fencing. Determine the equation of the curve of best fit, and use the equation to determine the value of x that minimizes the area of each square.
(Must use quadratics, no derivatives/calculus)
Only thing I've seen to do is A+=+X%5E2+%2B+Y%5E2 then 100+=+4x+%2B+4y. 25=x%2By, y=25-x.
A=X%5E2+%2B+%2825-x%29%5E2
A=x%5E2+%2B+625+-+25x+-+25x+%2B+x%5E2 ***************
A=x%5E2+-+50x+%2B+625 **************** A+=+2x%5E2+-+50x+%2B+625
But then solutions for x are imaginary, what do I do now?
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A=2x%5E2+-+50x+%2B+625
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There are no real zeroes for x, but that's not relevant.
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A=2x%5E2+-+50x+%2B+625
The minimum of the parabola is the vertex @ x = -b/2a
x = -50/-4 = 12.5 --> minimum Area
Minimum area = 312.5 sq m