SOLUTION: The 200m fencing is used to form a triangular plot in the corner of the allotments with two walls being used as two sides (right angle triangle).
As before, you experiment with di
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Question 975697: The 200m fencing is used to form a triangular plot in the corner of the allotments with two walls being used as two sides (right angle triangle).
As before, you experiment with different dimensions until you find the maximum area which can be enclosed.
Your results are based on 200m as stated above but you have developed a formula for each situation so that the area can be calculated whatever the length of fencing used.
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
x and y for the legs, h for the hypotenuse.
A, for area.
-
All that just from the first, "200" perimeter equation. This means the two formulas for h^2 can be equated to eliminate h.
Both sides have x^2+y^2...
-----can this be solved either for x or for y?
Use THAT formula of y in the area equation:
-----formula for AREA, in terms of only x, according the the description's conditions.
----Not displaying; not sure why.
You can use Google search engine, input y=x((200x-20000)/(x-200)) and enter, and you can see the graph, and can zoom in and shift the position of the framing to see a good local maximum. "y" as used in the graphing tool will stand for AREA, not the y variable of the dimension as in the problem description.
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