# Linear algebra question

• Jan 2nd 2013, 06:45 PM
elmidge
Linear algebra question
Hi. I thing I'm missing something simple, but can anyone help me solve the following:

A role of fence wire 1200m long is to be used to enclose and seperate two identical adjoining rectangular paddocks (which share one side-width).

Find the dimensions of the enclosure, so the total area is maximised.

Thanks in advance for the help.
• Jan 2nd 2013, 07:03 PM
Deveno
Re: Linear algebra question
A = WL

2W + 3L = 1200

W = 600 - 3L/2

A(L) = (600 - 3L/2)L = 600L - 3L2/2

take the derivative of A with respect to L (dA/dL) and set it equal to 0, then solve for L.
• Jan 2nd 2013, 07:57 PM
Soroban
Re: Linear algebra question
Hello, elmidge!

Did you make a sketch?

Quote:

A roll of fence wire 1200m long is to be used to enclose and seperate
two identical adjoining rectangular paddocks (which share one side-width).

Find the dimensions of the enclosure so the total area is maximised.

Let $\displaystyle x$ = total length of the paddocks.
Let $\displaystyle y$ = width of the paddocks.
Code:

      : - - - - - x - - - - - :       *-----------*-----------*       |          |          |       |          |          |     y|          |y          |y       |          |          |       |          |          |       *-----------*-----------*       : - - - - - x - - - - - :
We have: .$\displaystyle 2x + 3y \:=\:1200 \quad\Rightarrow\quad y \:=\:400 - \tfrac{2}{3}x$ .[1]

The area is: .$\displaystyle A \:=\:x\cdot y$

Substitute [1]: .$\displaystyle A \:=\:x\left(400-\tfrac{2}{3}x\right) \quad\Rightarrow\quad A \:=\:-\tfrac{2}{3}x^2 + 400x$

This is a down-opening parabola: $\displaystyle \cap$
Its maximum is at its vertex: .$\displaystyle x \:=\:\tfrac{\text{-}b}{2a}$

Hence: .$\displaystyle x \:=\:\frac{\text{-}400}{2(\text{-}\frac{2}{3})} \quad\Rightarrow\quad \boxed{x\:=\:300\text{ m}}$

Substitute into [1]: .$\displaystyle y \:=\:400 - \tfrac{2}{3}(300) \quad\Rightarrow\quad \boxed{y \:=\:200\text{ m}}$