Finding the range for the length of one side

[ndogspaintball88]

"A rectangular area is fenced in with 130 feet of fence. If the minimum area should be 1,000 square feet, what is the range of feet allowed for the length of the rectangle?"

I understand that this needs to be set up as an inequality in some way since it says the minimum area and it wants a range. Please help me to get this problem started, I do not know how to set it up

 

[soroban]

Hello, ndogspaintball88!

A rectangular area is fenced in with 130 feet of fence.
If the minimum area should be 1,000 square feet,
what is the range of feet allowed for the length of the rectangle?

              L
      * - - - - - - - *
      |               |
      |               |
    W |               | W
      |               |
      |               |
      * - - - - - - - *
              L

The perimeter is 130: .\(\displaystyle 2L + 2W \:=\:130 \quad\Rightarrow\quad W \:=\:65-L\) .[1]

The area is at least 1000 square feet: .\(\displaystyle A \;=\;LW \:\geq \:1000\) .[2]

Substitute [1] into [2]: .\(\displaystyle L(65-L) \:\geq \:1000\)

. . This simplifies to: .\(\displaystyle L^2 - 65L + 1000 \:\leq \:0\)


This is a up-opening parabola: .\(\displaystyle y \:=\:x^2 - 65x + 1000\)
When is it negative? .When is its graph below the x-axis?
. . Answer: between its x-intercepts.

\(\displaystyle x^2-65x + 1000 \:=\:0 \quad\Rightarrow\quad (x - 25)(x - 40) \:=\:0 \quad\Rightarrow\quad x \:=\:25,\:40\)


Therefore, the range of the length is: .\(\displaystyle 25 \:\leq \:L \:\leq \:40\)

 

[Denis]

If L > W and you want integers, then L = 33 to 40.
High: 33 by 32 = 1056
Low: 40 by 25 = 1000