1. 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.

2. 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.

3. Re: Linear algebra question

Hello, elmidge!

Did you make a sketch?

A roll of fence wire 1200m long is to be used to enclose and seperate

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

Let $x$ = total length of the paddocks.
Let $y$ = width of the paddocks.
Code:
      : - - - - - x - - - - - :
*-----------*-----------*
|           |           |
|           |           |
y|           |y          |y
|           |           |
|           |           |
*-----------*-----------*
: - - - - - x - - - - - :
We have: . $2x + 3y \:=\:1200 \quad\Rightarrow\quad y \:=\:400 - \tfrac{2}{3}x$ .[1]

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

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

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

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

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