SOLUTION: A rectangular plot of land is to be enclosed by fencing. One side is along a river and so needs no fence. If the total fencing available is 1400 meters, find the dimensions of the

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Question 1154933: A rectangular plot of land is to be enclosed by fencing. One side is along a river and so needs no fence. If the total fencing available is 1400 meters, find the dimensions of the plot to have maximum area. (Assume that the length is greater than or equal to the width.)
Answer by ikleyn(52798)   (Show Source): You can put this solution on YOUR website!
.

Since one side is the river, the rectangle's fence perimeter will be

L + 2W = 1400.

Hence, L = 1400 - 2W.

Area = Length * Width.

Substitute (1400-2W) for L:

A = W(1400 - 2W)

A = -2W^2 + 1400W.

This is a quadratic function. It has the maximum at x = -b/(2a), according to the general theory.

    (See the lessons
     
         - HOW TO complete the square to find the minimum/maximum of a quadratic function

         - Briefly on finding the minimum/maximum of a quadratic function

     in this site).


For our quadratic function the maximum is at

W =  =  = 350.

So, W = 350 ft is the width for max area.

Then the length is  L = 1400 - 2W = 1400 - 2*350 = 700 ft.


Find the max area. It is

A = L*W = 700*350 = 245000 square feet.


The plot of the quadratic function for the area is shown below:  y = area and x = width.


     


See my lessons in this site on finding the maximum/minimum of a quadratic function

    - HOW TO complete the square to find the minimum/maximum of a quadratic function
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola
    - A rectangle with a given perimeter which has the maximal area is a square
    - A farmer planning to fence a rectangular garden to enclose the maximal area
    - A rancher planning to fence two adjacent rectangular corrals to enclose the maximal area
    - Finding the maximum area of the window of a special form
    - Using quadratic functions to solve problems on maximizing revenue/profit


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