Word problem

[Timcago]

Problem: A racher who wishes to fence off a rectangular area finds that the fencing in the east west direction will require extra reinforcement owing to strong prevailing winds. Because of this, the cost of fencing in the east-west direction will be $12 per yard, as opposed to a cost of $8 per yard in the north-south direction. Fine the dimensions of the largest possible area that can be enclosed for $4800.

Here is how i attempt to answer it.

A = L*w
P=2L+2W

2(L+8)+2(W+12) --> 2L+16+2W+24 --> 2L+2W+40 --> 2L = -2W-40
--> L = -W-20

A= (-W-20)W --> A=-W^2-20W

A(x)= -(4800)^2-20(4800)

A(x) = 23040000 - 96000=22944000

This obviously is not right, can anyone help?

 

[soroban]

Hello, Timcago!

Sorry, I couldn't follow your work at all . . .
\(\displaystyle \;\;\)especially the way you stuck in the "4800".

A racnher who wishes to fence off a rectangular area
finds that the fencing in the east west direction will require extra reinforcement owing to strong prevailing winds.
Because of this, the cost of fencing in the east-west direction will be $12 per yard,
as opposed to a cost of $8 per yard in the north-south direction.
Find the dimensions of the largest possible area that can be enclosed for $4800.

Let \(\displaystyle L\) = length of the east-west fencing ($12/yard).
Let \(\displaystyle \,W\) = length of the north-south fencing ($8/yard).

There are two lengths \(\displaystyle L\) at $12/yard . . . cost: \(\displaystyle 24L\) dollars.
There are two widths \(\displaystyle W\) at $8/yard . . . cost: \(\displaystyle 16W\) dollars.

Since the total cost is limited to $4800: \(\displaystyle \,24L\,+\,16W\:=\:4800\;\;\Rightarrow\;\;W\:=\:300\,-\,\frac{3}{2}L\;\) [1]


We want to maximize the area: \(\displaystyle \,A\;=\;L\cdot W\)

Substitute [1]: \(\displaystyle \:A\;=\;L(300\,-\,\frac{3}{2}L)\;=\;300L\,-\,\frac{3}{2}L^2\)

Differentiate, equate to zero, and solve:
\(\displaystyle \;\;A'\;=\;300\,-\,3L\;=\;0\;\;\Rightarrow\;\;L\,=\,100\)

Substitute into [1]: \(\displaystyle \,W\:=\:300\,-\,\frac{3}{2}(100)\:=\:150\)


Therefore: use \(\displaystyle 100\) yards east-west and \(\displaystyle 150\) yards north-south.