# Complete the Square? Parabola?

• Feb 2nd 2010, 12:07 PM
Complete the Square? Parabola?
Alright, so I recently started my new semester, and I've forgotten almost everything that I've learned the previous year regarding math. I was assigned to do various questions, and I'm not exactly sure how to do it. The question is:

Rebecca has 600 m of fencing for her cornfield. The creek that goes through her farmland will form one side of the rectangular boundary. Rebecca considers different widths to maximize the area enclosed.

A. What are the minimum and maximum values of the width of the field?

I'm assuming that I will need to create a parabola for this question, due to the question asking a minimum and maximum. I also have the feeling completing the square will be involved, but I pretty much forgot how to do both. But here's what I got so far:

P = L + 2W (the creek makes up 1 length?)
L = 600 - 2W
A = W X L

That's all I got for now... any help, please?

B. What equations describe each?
i) The relationship between the length and width of the field.
ii) The relationship between the area and width of the field.

C. Copy and complete these table of values for widths that go from the least to greatest possible values in intervals of 50 m.

Width (m) | Length (m) | Area (m^2)
• Feb 2nd 2010, 12:23 PM
pickslides
Quote:

P = L + 2W (the creek makes up 1 length?)
L = 600 - 2W
A = W X L

That's all I got for now... any help, please?

You are doing very well up to here.

From your working, $\displaystyle A = W X L$ is really

$\displaystyle A = W \times (600 - 2W)$

$\displaystyle A = 600W - 2W^2$

Do you know how to find the maximum of this parabola? You need to make $\displaystyle A=0$ to solve for $\displaystyle W$ . The maximum value will be half way between the 2 solutions.
• Feb 2nd 2010, 12:36 PM
P = L + 2W
L = 600 - 2W
A = W X L
A = W X (600 - 2W)
A = 600W - 2W^2
A = -2W^2 + 600W
A = -2(W^2 - 300W) -> (300/2)^2 = 22500
A = -2(W^2 - 300W + 22500 - 22500)

Alright, got this so far. But I'm not sure what to do from here. From what I remember, I will have to use the -2 again on the next line, and isolate the -22500, but how do I do this... if that's what I'm even supposed to do?
• Feb 2nd 2010, 12:44 PM
pickslides
Quote:

P = L + 2W
L = 600 - 2W
A = W X L
A = W X (600 - 2W)
A = 600W-2W^2

You really don't need to complete the square from here, it would just be making life difficult and who has time for that?

To solve for $\displaystyle W$ make $\displaystyle A=0$

$\displaystyle 0 = 600W-2W^2$

Take common factors from the RHS

$\displaystyle 0= 2W(300-W)$

Using the null factor law $\displaystyle W = 0,300$

Quote:

Originally Posted by pickslides
The maximum value will be half way between the 2 solutions.

What is half way between 0 and 300? $\displaystyle W=\dots$

Put this value into $\displaystyle A = 600W-2W^2$ for $\displaystyle W$ to get max area.
• Feb 2nd 2010, 12:56 PM
P = L + 2W
L = 600 - 2W
A = W X L
A = W X (600 - 2W)
A = 600W - 2W^2
0 = 600W - 2W^2
0 = 2W(300 - W)

Wait, where did the 0,300 come from? In order for the right side of the equation to = 0, W would have to = 300.

Half way between 0 and 300 = 150. Therefore W (min value?) = 150.

A = 600W - 2W^2
A = 600(150) - 2(150)^2
A = 90000 - 2(22500)
A = 90000 - 45000
A = 45000

Therefore max area is 45000. Though, the question is asking for the minimum and maximum of the width of the field. Is the max area the max value for the width?
• Feb 2nd 2010, 01:02 PM
pickslides
Quote:

P
0 = 2W(300 - W)

Wait, where did the 0,300 come from? In order for the right side of the equation to = 0, W would have to = 300.

$\displaystyle 0 = 2W(300 - W) \implies 2W=0,300-W=0\implies W=0,300$

Quote:

Therefore max area is 45000. Though, the question is asking for the minimum and maximum of the width of the field. Is the max area the max value for the width?

Your answer for max width is 150 as $\displaystyle W=150$
• Feb 2nd 2010, 01:19 PM
Alright, thanks very much for the help. (Wink)

Before even attempting to complete the square, I guessed that the minimum width was 1, and the maximum width was 299. I guessed 1 because the length could be 598, and the two widths would be one each, equaling 600 (perimeter), but I guess that was wrong.

B. What equations describe each?
i) The relationship between the length and width of the field.

A = W X L

ii) The relationship between the area and width of the field.

W = A/L

These look correct?
• Feb 2nd 2010, 01:29 PM
pickslides
Quote:

These look correct?

They are both worng

Quote:

B. What equations describe each?
i) The relationship between the length and width of the field.

You need someithng with only $\displaystyle L$ and $\displaystyle W$ in it. I like

$\displaystyle L = 600 - 2W$

Quote:

ii) The relationship between the area and width of the field.

You need someithng with only $\displaystyle A$ and $\displaystyle W$ in it. I like

$\displaystyle A = 600W - 2W^2$
• Feb 2nd 2010, 01:40 PM
Got it now, ty.

C. Copy and complete these table of values for widths that go from the least to greatest possible values in intervals of 50 m.

Width (m) | Length (m) | Area (m^2)

Width (m)
0
50
100
150
200
250
300

Length (m)
600
500
400
300
200
100
0

Area (m^2)
0
25000
40000
45000
40000
25000
0

A friend shared these results with me. I don't even see any pattern within them at all. Are they correct? Will I have to somehow apply my results in questions A and B to complete this chart?
• Feb 2nd 2010, 01:53 PM
pickslides
I would suggest to used the information from the previous question

Width (m) | Length (m) | Area (m^2)

0 $\displaystyle ~~~~~~$ L = 600-2(0) $\displaystyle ~~~~~~$ A=600(0)-2(0)^2

50 $\displaystyle ~~~~~~$ L = 600-2(50) $\displaystyle ~~~~~~$ A=600(50)-2(50)^2

100 $\displaystyle ~~~~~~$ L = 600-2(100) $\displaystyle ~~~~~~$ A=600(100)-2(100)^2

etc....