# Thread: quadratic equations word problems geometry (math 20 pure)

1. ## quadratic equations word problems geometry (math 20 pure)

Hey! I'm having trouble with a certain question.

A rectangular field is to be enclosed by 600 m of fence.
a) What dimensions will give a maximum area?
b) What is the maximum area?

This is what I have so far:

x= 2 sides of rectangular field
y= 2 other sides of rectangular field

I know that:
2x+2y=600
and for the area I need to do (x)(y) = area.

Now what???

Thank you : )

2. Where's that function notation you studied earlier? You need it now.

Area is a function of x and y

Area(x,y) = x*y

We also know that x and y are related, so y can be expressed in terms of x.

2x + 2y = 600

x + y = 300

y = 300 - x

After substitution, this gives

Area(x) = x*(300-x) = 300x - x^2

This should look like a nice quadratic or parabola. You should have studied how to find the maximum of such things.

Show us what you get.

3. When I graphed " 300x - x^2 " in my calculator, it gave me a straight line.
Am I doing something wrong?

4. no, it is a parabola. i will look like a "line" in small intervals..

anyways, the point is, you have to find the maximum area right? so, could you show us what you did?

5. Haha, you were right, it did work on my calculator. I found the maximum on my calculator and it gave me x = 150 and y = 22500, but I have no idea how to use those results to answer the two questions that go along with the word problem.

6. Originally Posted by dancingqueen9
Haha, you were right, it did work on my calculator. I found the maximum on my calculator and it gave me x = 150 and y = 22500, but I have no idea how to use those results to answer the two questions that go along with the word problem.
take note of the $y=22500$. this is your $A(150)$ or the maximum area..

now, you have already have one dimension, what is the other dimension? use the constraints..