SOLUTION: A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle. Given that the total length of the fence is 80m show that the area, A, of the

Algebra ->  Surface-area -> SOLUTION: A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle. Given that the total length of the fence is 80m show that the area, A, of the      Log On


   

Question 1074737: A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle.
Given that the total length of the fence is 80m show that the area, A, of the garden is given by the formula A=y(80-2y), where you is the distance from the house to the end of the garden.
Given that the area is a maximum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
y, length from end of garden to house

Two of the sides sum to y%2By=2y. The side opposite the house must finish the length of the fence material of 80; so 80-2y is this length.

The two dimensions are y and 80-2y.
Area A=y%2880-2y%29.

Maximum area is when the rectangle becomes square shaped.
Since only three sides are to be used for the fence around the garden, each side is 80%2F3=26%262%2F3 meters.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
The solution by "josgarithmetic" is   highlight%28WRONG%29.

Below find the correct solution. 

The area of the garden, under the given condition, is

A = y*(80-2y) = -2y^2 + 80y.


Referring to the general form of a quadratic function

A = ay^2 + by +c,

the function have a maximum at y = -b%2F%282a%29, which is y = -80%2F%282%2A%28-2%29%29 = 90%2F4 = 20.


Thus the maximum is achieved at y = 20 m, and the maximal value of the quadratic function (of the area) is 

A = -2*20^2 + 80*20 = -2*400 + 1600 = 800 square meters.


Answer.  The dimensions of the garden are 20 m x 40 m. Its area is 800 square meters.


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The statement 
     At given perimeter, the area of a rectangle is maximal if and only if the rectangle is a square

is true if the PERIMETER (the entire perimeter consisting of four sides) is constrained.

In the given problem we have ANOTHER/DIFFERENT situation. (with which "josgarithmetic" is unfamiliar, due to his mathematical illiteracy).
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Plot A = -2y%5E2+%2B+80y


On finding the maximum/minimum of a quadratic function 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
    - 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 farmer planning to fence a rectangular area along the river to enclose the maximal area (*)
    - A rancher planning to fence two adjacent rectangular corrals to enclose the maximal area
    - Using quadratic functions to solve problems on maximizing revenue/profit
in this site.


Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic "Finding minimum/maximum of quadratic functions".


In the list of lessons, one is marked by the (*) sign.
It is your prototype/sample.


        H a p p y   l e a r n i n g  ! !