SOLUTION: a farmer needs to enclose a rectangular field with a fence down the middle to divide it into two equal parts. The material for the divider down the middle cost $50 per foot ant the

Algebra ->  Rectangles -> SOLUTION: a farmer needs to enclose a rectangular field with a fence down the middle to divide it into two equal parts. The material for the divider down the middle cost $50 per foot ant the      Log On


   

Question 1159033: a farmer needs to enclose a rectangular field with a fence down the middle to divide it into two equal parts. The material for the divider down the middle cost $50 per foot ant the material for the outside cost $20 per foot. He wants the area of the field to be 1600 sq/ft. what dimensions will minimize the cost of material
write an expression involving x and y giving the total cost C of fencing materials
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

From the context, the dimensions of the rectangle are not given for advance - they are unknowns and they should be found

from the minimum cost condition.


Let x be one dimension and y be the other dimension of the rectangle.


Then the cost of the outside perimeter fence is 20*(2x+2y) dollars = 40*(x+y) dollars,

while the cost of the fence down middle is 50*x dollars.


The total cost of the fence is  

    C(x,y) = 40x + 40y + 50x = 90x + 40y  dollars.      ANSWER


It shoud be minimized under the restriction

    x*y = 1600  square feet.

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The problem's question is just answered; thus the problem's solution is completed.

For a TWIN problem, the full solution,  including minimization part,  is in the lesson
    - Calculus optimization problems for shapes in 2D plane, Problem 5
in this site.


Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!


Cost = $20∙(AB+BC+CD+DE+EF+FA) + $50∙(BE)

Cost = 20∙(x+x+h+x+x+h) + 50∙(h)

Cost = 20(4x+2h) + 50h

Cost = 80x+40h+50h

Cost = 80x+90h

Area = AC∙CD = (2x)∙(h) = 1600

2xh = 1600

h = 1600/(2x) = 800/x

Cost = C = 80x+90(800/x)



Set the derivative equal to 0, to find potential extrema:



Now we test to see if x=30 produces the minimum cost.
We first try the 2nd derivative test



When we substitute 30 for x, we will get a positive number,
so the curve is concave upward at x=30, so x=30 produces
the minimum cost.

So the dimensions are when x=30 ft. and h = 800/x = 800/30 = 26 2/3 ft.
But since x is only half the base, the actual dimensions are 60 ft by 26 2/3 ft.

The minimum cost is Cost = C = 80(30)+7200/30 = $2640.

Edwin