SOLUTION: Sabrina is building a garden against the back wall of her house. She needs to put mesh fence around the three outside edges to stop the racoons getting into the garden. If she has

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Question 1195548: Sabrina is building a garden against the back wall of her house. She needs to put mesh fence
around the three outside edges to stop the racoons getting into the garden. If she has 60 m of
mesh fence, what dimensions will optimize the area of the garden? What is the maximum area
the garden?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
Sabrina is building a garden against the back wall of her house. She needs to put mesh fence
around the three outside edges to stop the racoons getting into the garden. If she has 60 m of
mesh fence, what dimensions will optimize the area of the garden? What is the maximum area
the garden?
~~~~~~~~~~~~~~~ 


Since one side is back wall of the house, the rectangle's fence perimeter will be

L + 2W = 60 meters.

where L is the dimension along the wall and W is the dimension perpendicular to the wall.
Hence, L = 60 - 2W meters.


    Area = Length * Width.


Substitute (60-2W) for L:

    A = W(60 - 2W)       (1)

    A = -2W^2 + 60W.


It is a quadratic function. It has the maximum at x = -b/(2a), where "a"  is the coefficient at the quadratic term 
and  "b"  is the coefficient at the linear term, 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).


In your case, the maximum is at


    W = -60%2F%282%2A%28-2%29%29 = %28-60%29%2F%28-4%29 = 15.


So,  W = 15 meters is the dimension perpendicular to the wall of the house. 


Then the length is  L = 60 - 2W = 60 - 2*15 = 30 meters.


Find the max area. It is 

    A = L*W = 30*15 = 450 square meters.    It is the maximum area.


ANSWER.  The dimensions are 30 m along the wall and 15 m perpendicular to the wall

         The maximum area is 450 m^2.

Solved.

----------------

My other lessons in this site on finding the maximum/minimum of a quadratic function are
    - 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



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


This problem has a somewhat surprising general solution:

The area is maximized when the side parallel to the wall uses half of the fencing.

If you encounter this kind of problem often (like the high school math team students I work with a lot), you can use that general result to get the answer to any problem like this quickly.

In this problem, with 60m of fencing, the length of fence parallel to the wall is 30m, leaving (60-30)/2 = 15m for the lengths of the other two sections of fence. That gives you the answer that the maximum area as 30*15 = 450 square meters.

This result is easy to obtain algebraically, using the methods shown in the response from the other tutor for finding the maximum value of a quadratic expression.

Let the total length of fencing be A; let the length of fence parallel to the existing wall be x. Then the width (the length of each of the two sections of fence perpendicular to the wall) is (A-x)/2. The area is then

A=x%28%28A-x%29%2F2%29
A=%28-1%2F2%29x%5E2%2B%28A%2F2%29x

The maximum value of this quadratic expression is when "x=-b/(2a)":

x=%28%28-A%2F2%29%2F%282%28-1%2F2%29%29%29=%28-A%2F2%29%2F%28-1%29=A%2F2

The general result even still holds if the area is divided into sections with multiple lengths of fence perpendicular to the wall.

If the amount of fencing is again A and the length of the fence parallel to the wall is x, and if there are n lengths of fence perpendicular to the wall (dividing the area into (n-1) sections), then the length of each section of fence perpendicular to the wall is (A-x)/n, and the area is

A=x%28%28A-x%29%2Fn%29
A=%28-1%2Fn%29x%5E2%2B%28A%2Fn%29x

The area is again maximum when the length x is half the total amount of fencing:

x=%28-A%2Fn%29%2F%28-2%2Fn%29=A%2F2

So, for example, if there are the same 60m of fencing, and the area is divided into 4 sections (using 5 lengths of fence perpendicular to the wall), then the area is maximum when the length is 60/2=30m and the width is (60-30)/5=6m; and the maximum area is 30*6=180 square meters.