SOLUTION: A rectangular area of 1,050 square feet is to be enclosed by a fence, then divided down the middle by another piece of fence. The fence down the middle costs $0.50 per running foot

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Question 1150956: A rectangular area of 1,050 square feet is to be enclosed by a fence, then divided down the middle by another piece of fence. The fence down the middle costs $0.50 per running foot, and the other fence costs $1.50 per running foot. Find the minimum cost for the required fence.
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
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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 1.50*(2x+2y) dollars = 3*(x+y) dollars,

while the cost of the fence down middle is 0.50*x dollars.


    Note, that I don't know now, which dimension will be the length and which be the width.

    When the problem will be solved, the solution will tell me it . . . 


So, I need minimize the function  

        f(x,y) = 3*(x+y) + 0.5x     (1)

under the condition

        x*y = 1050.                 (2)



From (2), express  y = 1050%2Fx  and substitute it into (1). You will get

        g(x) = 3%2A%28x%2B+1050%2Fx%29+%2B+0.5x = 3x + 3150%2Fx + 0.5x.


Differentiate it over x

        g'(x) = 3 - 3150%2Fx%5E2 + 0.5

and equate the derivative to zero.  You will get


        3.5 = 3150%2Fx%5E2,    or

        3.5x^2 = 3150

           x^2 = 3150%2F3.5 = 900,

           x   = sqrt%28900%29 = 30.


Thus the dimensions of the rectangle are 30 ft  and  1050%2F30 = 35 ft.

The fence down middle has the length of 30 ft;  hence, it is parallel to the shorter side of the rectangle.

Solved.

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If you want to see many other similar solved problems, look into the lesson
    - Calculus optimization problems
in this site.