Optimization Example: Area of Rectangle

We have a string of length \ell. It does not stretch. But we want to form a rectangle using this string as its boundary. Moreover we want this rectangle’s area is maximized. What to do? (Spoiler alert: solution below!)

Well, if we form a rectangle, then it has sides h and w satisfying 2w+2h=\ell (the perimiter is equal to the length of string).

The area would be
(1)\displaystyle A(w,h)=w\cdot h
This seems simple enough. We impose our constraint to write
(2)\displaystyle h = \frac{\ell}{2}-w
Thus our area function becomes
(3)\displaystyle A(w)=w\left(\frac{\ell}{2}-w\right).
How do we find the optimal area?

Take its derivative! We find
(4)\displaystyle A'(w)=\frac{\ell}{2}-2w.
Setting this to zero gives us
(4′)\displaystyle A'(w)=\frac{\ell}{2}-2w=0.
Its solution is
(5)\displaystyle \frac{\ell}{4}=w.
Plugging this back into (2) to find the value of h produces
(6)\displaystyle h = \frac{\ell}{4}.
Thus the sides to the rectangle are equal!

So the optimal rectangle (which maximizes its area with its perimiter fixed) is a square!


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Optimization. Bookmark the permalink.

2 Responses to Optimization Example: Area of Rectangle

  1. Pingback: [Index] Calculus of a Single Variable | My Math Blog

  2. Pingback: [Index] Calculus in a Single Variable | My Math Blog

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