A farmer wishes to build a large rectangular grazing area of dimensions, width (W) and length (L) splitting the area into 5 equal fields by dividing the width into 5 equal pieces. The total area of...
A farmer wishes to build a large rectangular grazing area of dimensions, width (W) and length (L) splitting the area into 5 equal fields by dividing the width into 5 equal pieces. The total area of all the fields must be 10500 square feet. Find the dimensions that use the least amount of fencing.
(It is not always the case that the width is shorter than the length. Round-off your answers to 3 decimal places)
The formula of area of rectangle is:
`A= Le n g t h* Width`
The given total area of the rectangular grazing lot is 10500 square feet. Its length is represented by L and width by W. So,
Solving for W yields:
`W= 10500/L ` (Let this be EQ1.)
Since the grazing lot is divided into five equal width, the fence includes the outside perimeter of the lot and the four boundaries inside. Note that the fence inside is parallel and equal to the length. Also, the total fence of the rectangular grazing lot refers to its total perimeter. So,
`P = 2L+2W+4L`
Express the total perimeter as one variable. To do so, substitute EQ1.
`P=6L + 21000/L`
To determine the dimensions that would minimize the perimeter of the rectangle, take the derivative of P with respect to L.
`P' = 6-21000/L^2`
Set P' equal to zero.
Isolate L. So move `21000/L^2` to the left side.
Multiply both sides by` L^2` .
Divide both sides by 6.
Take the square root of both sides.
Since L represents the length of the rectangle, take only the positve value. So,
Substitute this value of L to EQ1.
Expressing the L and W to its equivalent decimal value yields:
`L = sqrt3500 = 59.16079783 ...`
`W=10500/sqrt3500= 177.4823934 ...`
Rounding off to three decimal places result to:
Hence the dimension of the rectangular grazing lot is `59.161 xx 177.482` feet.
We have that the perimeter of the field is given by
`P= 6l +2w`
and that the area is given by
`A = lw = 10500`
Notice the symmetry of the problem. The perimeter will be maximised when
`6l = 2w`
`implies` `w = 3l`
Therefore we have that
`A = 3l^2 = 10500`
`implies` `l = sqrt(10500/3) = 59.161` ft
Then `w = 3l = 3(59.16) = 177.482` ft
The length of the field should be 59.161 ft
The total width of the field should be 177.482 ft