SOLUTION: This is a math stars question I can't don't know how to tell my son how to get the answer. Faye has 20 feet of fencing to make a rectangular pen for her dog. what is the largest a

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Question 165454: This is a math stars question I can't don't know how to tell my son how to get the answer.
Faye has 20 feet of fencing to make a rectangular pen for her dog. what is the largest area that she can fence in? His teacher said answer is 25 square feet-how did she get that answer?

Found 2 solutions by Alan3354, scott8148:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
This is a math stars question I can't don't know how to tell my son how to get the answer.
Faye has 20 feet of fencing to make a rectangular pen for her dog. what is the largest area that she can fence in? His teacher said answer is 25 square feet-how did she get that answer?
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L = length
W = width
2L + 2W = 20, L + W = 10, or W = 10 - L
Area = L*W = L*(10-L)
A = 10L - L^2
For maximum, set the 1st derivative equal to zero.
10 - 2L = 0
L = 5 (for max area)
W is also 5, so the max area is 25 sq ft.


Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
the largest area enclosed by a rectangle is actually a square
__ a square with a perimeter of 20ft has four 5ft sides
__ so the area is 25sqft


to get technical __ the perimeter is 20ft, so L+W=10 (half the perimeter) __ L=10-W

the area is L times W __ A=L*W __ substituting A=(10-W)*W __ A=10W-W^2

A=-W^2+10W is a downward opening parabola __ the maximum value for A occurs at the axis of symmetry

the equation for the axis of symmetry is W=-(10)/(2*(-1)) __ W=5

substituting __ L+(5)=10 __ L=5

A=L*W __ A=5*5 __ A=25