Question 1117428
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<pre>
Let the enclosure dimensions be  x = length, y = width, so the area is xy = 4000 square meters.


Then you have a rectangle of the perimeter of 2x + 2y plus two additional fence parts of the length y. 


The total length of the fencing with two additional fence parts is 2x + 4y meters.


Thus you need to minimize  2x + 4y  under the condition  xy = 4000.


Express y = 4000/x  and substitute it into  2x + 4y.  Then you will get that the function to minimize is  {{{2x + 4*(4000/x)}}}.


To find a minimum, differentiate the function over x and equate to zero. 


So, the equation for the minimum is


{{{2}}} - {{{16000/x^2}}} = 0,   or  {{{16000/x^2}}} = 2  ====>  {{{x^2}}} = {{{16000/2}}} = 8000  ====>  x = {{{sqrt(8000)}}} = {{{40*sqrt(5)}}}.


Thus the optimal dimensions are: the length = {{{40*sqrt(5)}}} = 89.44 m  and

                                 the width = {{{4000/(40*sqrt(5))}}} = {{{(100*sqrt(5))/5}}} = {{{20*sqrt(5)}}} = 44.72 m.


<U>Check</U>.  89.44*44.72 = 4000 square meters.   ! Correct !
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Solved.