SOLUTION: problem solving using quadratic functions: Assume the perimeter of a rectangle is P and that the area is to be maximized. Let x represent the width of the rectangle. Using quadra

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: problem solving using quadratic functions: Assume the perimeter of a rectangle is P and that the area is to be maximized. Let x represent the width of the rectangle. Using quadra      Log On


   

Question 1122448: problem solving using quadratic functions:
Assume the perimeter of a rectangle is P and that the area is to be maximized. Let x represent the width of the rectangle. Using quadratic functions, show that the rectangle must be a square.
Found 2 solutions by ikleyn, htmentor:
Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!

A rectangle with a given perimeter which has the maximal area is a square


There are many rectangles with a given perimeter.
For example, if the perimeter is  200 feet,  you can consider rectangles  20 by 80  feet,  30 by 70  feet,  40 by 60 feet,  50 by 50  feet and many others.

The amazing fact is:  among rectangles with a given perimeter,  a square has the maximal area.

Theorem

Among rectangles with a given perimeter,  a square has the maximal area.

Proof 

Let P be the given perimeter of a rectangle, and let L be the length and W be the width of the rectangle.

Then we have 

2L + 2W = P,

so we can express W as

W = %28P-2L%29%2F2.

The area of the rectangle is 

L*W = %28L%2A%28P-2L%29%29%2F2 = -L%5E2+%2B+%281%2F2%29%2APL.




  

    Now let me remind you that, if you have a quadratic function f(x) = ax%5E2+%2B+bx+%2B+c of the general form,   
    then it reaches the maximum/minimum at x = -b%2F2a.
  



      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 situation,  L = %28%28P%2F2%29%29%2F%28%282%29%29 = P%2F4.

Then W = %28P-2L%29%2F2 = %28P-2%2A%28P%2F4%29%29%2F2 = P%2F4,

and the rectangle is the square with L = W =  P%2F4.

The statement is proved.

Let me repeat the statement one more time:

    Among rectangles with a given perimeter  P,  a square with the side length   P%2F4   has the maximal area,  which is equal to   %28P%2F4%29%5E2.    

Example 1

A rectangle has a perimeter of  100 meters.  What are the dimensions of the sides if the area is a maximum?

Solution

According to the proved statement above,  the rectangle is a square with the side length of  100%2F4 = 25 meters.


Problem 1

A piece of wire  20 inches long is to be cut into a rectangle.  How long should each side of the rectangle be to maximize the area of the rectangle?

Solution

According to the proved statement above,  the rectangle should be a square with the side length of  20%2F4 = 5 inches.


Problem 2

A farmer has  280 feet of fence to enclose a rectangular garden.  What dimensions for the garden give the maximum area?

Solution

According to the proved statement above,  the rectangular garden should be a square with the side length of  280%2F4 = 70 feet.

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

What you see in this post, is the lesson
    - A rectangle with a given perimeter which has the maximal area is a square
from 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".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.



Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
P = 2x + 2y, where y = the length of the rectangle
The area, A = x*y
From the first equation, y = P/2 - x
Thus A = (P/2 - x)x
The area will be maximized if dA/dx = 0
dA/dx = P/2 - 2x = 0
This gives x = P/4
Thus y = P/2 - P/4 = P/4
x = y = P/4, which implies the rectangle must be a square