Thread: Quick word problem

A rectangular area is enclosed by a fence and divided by another section of fence parallel to two of its sides. If the 600m of fence used encloses a maximum area, what are the dimensions of the enclosure?

My work:

2L + 3W = 600
2L = 600 - 3W
L = 300 - 1.5W
A = LW
A = (300 - 1.5W)W
A = w(300 - 1.5W)
Zeros: X=0, X= 200
200/2 = 100
Width: 100
Length: 200

Correct ?

Hello, irishad!

Thank you for showing your work!

A silly arithmetical error at the very end . . .

A rectangular area is enclosed by a fence and divided
by another section of fence parallel to two of its sides.
If the 600m of fence used encloses a maximum area,
what are the dimensions of the enclosure?

My work:

$\displaystyle 2L + 3W \:= \:600\quad\Rightarrow\quad 2L \:= \:600 - 3W \quad\Rightarrow\quad L \:= \:300 - 1.5W$

$\displaystyle A \:= \:LW \;=\;(300 - 1.5W)W \;=\;W(300 - 1.5W)$

Zeros: .$\displaystyle x=0,\;x= 200\quad\Rightarrow\quad \frac{200}{2}\:=\:100$

.Width: .$\displaystyle 100$
Length: .$\displaystyle 200 \quad {\color{red}\leftarrow\quad oops!}$

The length is: .$\displaystyle L \,=\,150$

But your reasoning (and the rest of your work) is absolutely correct.

. . Good work!

How is the length 150 ? :S

Hello, irishad!

How is the length 150 ?

I have a question . . . Where in the world did you get 200 ?


You had:. $\displaystyle L \:=\:300 - 1.5W$

Then you found: .$\displaystyle W \,=\,100$

Therefore . . .