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jayglass16
06-19-2011, 12:33 PM
The question is:
A rectangular lot is to be fenced with 80m of fencing. What is the maximum area and what are the dimensions pf the lot?


Any ideas on how to solve?? Im Stumped

JeffM
06-19-2011, 02:05 PM
The question is:
A rectangular lot is to be fenced with 80m of fencing. What is the maximum area and what are the dimensions pf the lot?


Any ideas on how to solve?? Im Stumped
Hmm Have you studied differential calculus? Probably not, but let's make sure.

If you have not, this problem is a hard one. Let's take it in baby steps OK?

First thing in almost any problem that gives you even a bit of trouble is to "name" things with a symbol and set out the known relationships in terms of those symbols.

Let w = width of the lot.
Let d = the depth of the lot.
Let a = the area of the lot.
Let p = the perimeter of the lot.

That was the "naming" part. Now what do we know about the relationships among those symbols?

Using the symbols above, what is the perimeter of the lot?
What is the area of the lot?

renegade05
06-21-2011, 01:38 AM
You will need a graphing calculator if you don't know differential calculus as JeffM said... or some trail and error.

I never liked the wording of these questions though. At the maximum area you end up getting a square lot, not a rectangular lot. Kinda misleading..

soroban
06-21-2011, 09:30 AM
Hello, jayglass16!

Calculus is not required . . .


A rectangular lot is to be fenced with 80m of fencing.
What is the maximum area and what are the dimensions of the lot?


x
* - - - - - *
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y | | y
| |
* - - - - - *
x
Let x = length of the lot.
Let y = width of the lot.

\text{The perimeter is 80: }\;2x + 2y \:=\:80 \quad\Rightarrow\quad y \:=\:40 - x .[1]

\text{The area is given by: }\;A \:=\:xy .[2]


Substitute [1] into [2]:
. . A \:=\:x(40 - x) \quad\Rightarrow\quad A \:=\:-x^2 + 40x

\text{We have a down-opening parabola: }\:\cap
\text{Its maximum is at its }vertex\!: \;x \:=\:\tfrac{\text{-}b}{2a}
\text{Hence: }\:x \:=\:\frac{\text{-}40}{2(\text{-}1)} \quad\Rightarrow\quad \boxed{x\:=\:20}

Substitute into [1]:
. . y \:=\:40 - 20 \quad\Rightarrow\quad \boxed{y \:=\:20}


\text{Therefore: }\;\begin{Bmatrix}\text{Dimensions:}& 20\text{m}\times20\text{m} \\ \text{Max. area:} & 400\text{ m}^2 \end{Bmatrix}