Question 818760: Can someone help me with this question please :-)
Bonnie Wolansky has 100 ft of fencing material to enclose a rectangular exercise run for dog. One side of the run will border her house, so she will only need to fence three sides. What dimensions will give the enclosure the maximum area? What is the maximum area?
Thank you
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! 100 = 2w + L
L = 100 - 2w
a = wL
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a = wL
a = w(100 - 2w)
a = 100w - 2w^2
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a(w) = -2w^2 + 100w
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the above quadratic equation is in standard form, with a=-2, b=100, and c=0
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to solve the quadratic equation, plug this:
-2 100 0
into this: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic vertex is a maximum at ( w= 25, a(w)= 1250 )
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Answer:
the maximum area is 1250 sq.ft ( a(w) from the vertex )
w = 25 ft ( w from the vertex )
L = 50 ft ( calculated from above equation for L using w )
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