SOLUTION: Calculus: A fence must be built to enclose a rectangular area of 5000 ft^2. Fencing material costs $1 per foot for the two sides facing north and south and ​$2 per foot fo

Algebra ->  Statistics  -> Confidence-intervals -> SOLUTION: Calculus: A fence must be built to enclose a rectangular area of 5000 ft^2. Fencing material costs $1 per foot for the two sides facing north and south and ​$2 per foot fo      Log On


   



Question 1029100: Calculus:
A fence must be built to enclose a rectangular area of 5000 ft^2. Fencing material costs $1 per foot for the two sides facing north and south and ​$2 per foot for the other two sides.
Find the cost of the least expensive fence.

Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A fence must be built to enclose a rectangular area of 5000 ft^2. Fencing material costs $1 per foot for the two sides facing north and south and ​$2 per foot for the other two sides.
Find the cost of the least expensive fence.
--------------
Area:: w*h = 5000
-----
Cost = 2(h + 2w)
---------
Substitute for h::
C(w) = 2(5000/w + 2w)
-----
C(w) = 10,000/w + 4w
-----
Take the derivative:
C'(w) = 10000(-1/w^2) + 4
------
Solve:: -10000/w^2 = -4
-------
w^2 = 2500
width = 50 ft ; so cost is 2*($2)50 = $200
height = 5000/50 = 100 ft ; so cost is 2($1)100 = $200
======================
Total cost = $400.00
-----------------------
Cheers,
Stan H.


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
You are asked to minimize expense
Let +s+ = the length of one of the sides
that face north and south and costs $1 / ft
---------------
Let +A+ = area enclosed by fences
+A+=+s%2A%28+5000%2Fs+%29+
So, the sides are +s+ and +5000%2Fs+
------------------
Let +P+ = the perimeter of the area fenced in
The perimeter of the area is:
+P+=+2s+%2B+2%2A%28+5000%2Fs+%29+
-----------------------
Let +C+ = the cost of fencing in the area
+C+=+1%2A%28+2s+%29+%2B+2%2A2%2A%28+5000%2Fs+%29+
+C+=+2s+%2B+20000%2Fs+
+C1+=+2+-+20000+%2F+s%5E2+
Set the 1st derivative, +C1+=+0+
+0+=+2+-+20000+%2F+s%5E2+
+20000+%2F+s%5E2+=+2+
+s%5E2+=+10000++
+s+=+100+
and
+5000%2Fs+=+5000%2F100+
+5000%2Fs+=+50+
-----------------------
Plug these results back into original equation
+C+=+2s+%2B+20000%2Fs+
+C+=+2%2A100+%2B+20000%2F100+
+C+=+200+%2B+200+
+C+=+400+
The least expensive fence costs $400
---------------------------------
check answer:
+A+=+s%2A%28+5000%2Fs+%29+
+A+=+100%2A50+
+A+=+5000+
OK -here's the plot of Cost, +C+ and +s+
+graph%28+500%2C+500%2C+-30%2C+200%2C+-100%2C+600%2C+2x+%2B+20000%2Fx+%29+