If 400 feet of fencing is used to enclose a rectangular plot of land that borders a river, what is the maximum area that can be enclosed?
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OpenStudy (anonymous):
It's pretty east to start out.
Use the formulas:
A = w * h
P = 2h + w
where P is the perimeter of the fence (400 ft), h is the length of one of the 2 parallel sides, and w is the length of the side that is parallel to the river.
OpenStudy (anonymous):
okay so 400 = 2h + w
now what?
OpenStudy (anonymous):
system of equations?
OpenStudy (anonymous):
so solve for either h or w in that equation and substitute that into the Area formula
OpenStudy (anonymous):
okay so w = 200/h
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OpenStudy (anonymous):
no quite
OpenStudy (anonymous):
so 400 = 2h + 200/h
OpenStudy (anonymous):
w = 400 - 2h
OpenStudy (anonymous):
argh okay ...
OpenStudy (anonymous):
so 400 = 2h + (400 - 2h)
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OpenStudy (anonymous):
so that w should be substituted into the Area formula:
A = w * h
so
A = (400 - 2h) * h
OpenStudy (anonymous):
A = 400 -2h^2
OpenStudy (anonymous):
A = 400h - 2h^2
OpenStudy (anonymous):
and now just find the derivative and set if equal to 0?
OpenStudy (anonymous):
and yes. take the derivative, set to 0, and that will give you h. you can find w from that, which will give you the total area
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OpenStudy (anonymous):
so then h = 100 right?
OpenStudy (anonymous):
indeed it does!
OpenStudy (anonymous):
is w just the x coordinate at x =100?
OpenStudy (anonymous):
i mean y coordinate?
OpenStudy (anonymous):
sure. if you're mapping w (y-coord) against h (x-coord).
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