# Thread: Simultaneous Equations: Area of a Rectangle

1. ## Simultaneous Equations: Area of a Rectangle

Hello everyone!

The textbook out of which I am studying provides examples of simultaneous equation problems, but nothing similar to what is being required here. Does anyone know what to do in parts c and d? I have drawn a diagram and done a and b, but I cannot solve the latter two. Much help is appreciated! Thank you in advance to any mathematicians out there.

A rectangular enclosure is to be fenced so that its perimeter is 60 metres.

a. If w is the length of the rectangle, what is its length in terms of w?
l = 30 - w

b. Write an expression for A, the area of the rectangular enclosure.
A = w(30 - w)

c. What value of w will give the greatest area of the rectangular enclosure?

d. What is the greatest area the rectangular enclosure can have?

2. Do you know how to find the derivative of A with respect to w?

3. No, I'm afraid not. I am only in Year 9. Is there any other possible way?

4. I agree with your a) and b).

You have a Quadratic equation for the area. So you should be able to put it in turning point form, so you can read off the maximum.

$A = w(30 - w)$

$A = 30w - w^2$

$A = -(w^2 - 30w)$

$A = -\left[w^2 - 30w + (-15)^2 - (-15)^2\right]$

$A = -\left[(w - 15)^2 - 225\right]$

$A = -(w - 15)^2 + 225$.

Therefore the turning point is $(15, 225)$. Since this is a "negative" quadratic, the turning point is a maximum. Therefore, the maximum area is $225\,\textrm{m}^2$ and occurs when the width is $15\,\textrm{m}$.

5. Note: Prove It used "completing the square".