Another Word Problem: dimensions of max. area

[iRishad]

A rectangular area is enclosed by a fence and divided by another section of fence parallel to two of its sides. If the 600m of fence used encloses a maximum area, what are the dimensions of the enclosure?

My work:

2L + 3W = 600
2L = 600 - 3W
L = 300 - 1.5W
A = LW
A = (300 - 1.5W)W
A = w(300 - 1.5W)
Zeros: X=0, X= 200
200/2 = 100
Width: 100
Length: 200

Correct ?

 

[tkhunny]

Re: Another Word Problem

A rectangular area is enclosed by a fence and divided by another section of fence parallel to two of its sides. If the 600m of fence used encloses a maximum area, what are the dimensions of the enclosure?

My work:

2L + 3W = 600
2L = 600 - 3W
L = 300 - 1.5W
A = LW
A = (300 - 1.5W)W

Perfect down to here. I would have used (3/2), rather than 1.5, but that might be just a personal preference. Maybe not. Give it some thought.

A = w(300 - 1.5W)

Why did the variable change to lower case?

Zeros: X=0, X= 200

Why did the variable change to 'X' and how did you get these? Some intermediate steps are missing.

200/2 = 100
Width: 100
Length: 200

You did not explain what you are doing, here. Where did your variables go? What's the point of defining variables if you just discard them half way through the otherwise very nice solution process?

 

[iRishad]

Re: Another Word Problem

A = w(300 - 1.5W)
Why did the variable change to lower case?
That didn't matter really, lol.

Zeros: X=0, X= 200
Why did the variable change to 'X' and how did you get these? Some intermediate steps are missing.
Well those are my 'zeros'/x-intercepts(on a parabola)

200/2 = 100
Width: 100
Length: 200
You did not explain what you are doing, here. Where did your variables go? What's the point of defining variables if you just discard them half way through the otherwise very nice solution process?
200/2 is for the -b/2(a) thing. So I can bascially get the optimum value.

I just wish to know if my answers(Width: 100, Length: 200) are correct or not. Thank you.

 

[Mrspi]

Re: Another Word Problem

A = w(300 - 1.5W)
Why did the variable change to lower case?
That didn't matter really, lol.
Actually, it DOES matter. "w" and "W" are two different variables. Sloppy notation invariably leads to mistakes. And to someone trying to analyze your work, it sure could make a big difference.

Zeros: X=0, X= 200
Why did the variable change to 'X' and how did you get these? Some intermediate steps are missing.
Well those are my 'zeros'/x-intercepts(on a parabola)
Your original equation does not involve x, so it is difficult for an outside observer to determine "what happened" here. And there's nothing wrong with having w-intercepts, if you've used w as the independent variable in your function.


200/2 = 100
Width: 100
Length: 200
You did not explain what you are doing, here. Where did your variables go? What's the point of defining variables if you just discard them half way through the otherwise very nice solution process?
200/2 is for the -b/2(a) thing. So I can bascially get the optimum value.
I really don't think you're doing the -b/2a thing here, since you haven't shown a quadratic in standard form ANYWHERE in your process. Maybe.....you are using the fact that the vertex of this parabola has a w-coordinate that is halfway between the two intercepts? And how did you get that length?

I just wish to know if my answers(Width: 100, Length: 200) are correct or not. Thank you.

Your answer is correct. We are just wishing you were a bit more careful in showing your work. And while you may think this does not matter since the answer is correct, please know that PROCESS is important.

 

[iRishad]

Re: Another Word Problem

Yeah sorry about that. I'm just in grade 8, so...we're not in depth I guess.

 

[tkhunny]

Re: Another Word Problem

"I just wish to know if my answers...are correct." Sorry, I really can't get behind that. You should explain and understand what you are doing. 8th grade or 27th grade, be careful and precise. You cannot be either of those if you do not write down what you are doing.

Variables changing IS a problem. It makes your presentation incomprehensible. I would take points off for it on an exam.

Explanations ARE important. I know what you are doing, but again, if I am grading your exam, there is no way for me to know that YOU know what you are doing unless you tell me.

Oddly, you SAID you were using -b/(2a), but you didn't really do that. You used the midpoint of the zeros.

You are also missing an important point. Not all coordiante axes are in x and y. In this case, you have Area(A) as a function of Width(W). This constitutes a set W-A Coordinate axes. There should be no x popping up.

Clean and Complete...

Define the Variables. (I missed this on the first pass. You didn't do it. WRITE DOWN your definitions.)

L is the Length of the rectangular area.
W is the Width of the rectangular area. This is the side with the parallel fence inside the area

The Total Fence Length
2L + 3W = 600

The Total Enclosed Area
A = LW

Using the Length formula to reduce the Area formula to a single variable.
2L = 600 - 3W
L = 300 - 1.5W
A = L*W = (300 - 1.5W)W
A = W(300 - 1.5W)

We recognize the graph of A(W) as a parabola opening down, since the coefficient on the $\displaystyle W^{2}$ term is negative. We can then use the zeros to find the maximum value.

0 = W(300 - 1.5W) ==> W = 0 and W = 200

The midpoint on the W-axis is the W-coordinate of the maximum value.

$\displaystyle W_{max}$ = (200+0)/2 = 100
Note: We could have found this directly, without finding the zeros, using the standard formula -b/(2a), with a and b being the standard assignment of the coefficients of the quadratic in W, giving $\displaystyle W_{max}$ = -300/(2*(-3/2)) = -300/(-3) = 100.

Using $\displaystyle W_{max}$ to evaluate $\displaystyle L_{max}$

$\displaystyle L_{max}$ = 300 - 1.5$\displaystyle W_{max}$ ==> $\displaystyle L_{max}$ = 200

This sort of presentation does several things:

1) It is complete and thorough.
2) You should get and you will deserve full credit.
3) Anyone grading the exam can see exactly what you have done and that you have a complete understanding of the concepts involved.
4) You will have no problem understanding what you have done when you look at it next week.
5) It may seem a little wordy and excessive, but in this way you can write your own textbooks. This is very helpful if you don't particularly care for your author's presentation.
6) It gives you utter and complete confidence in your result. You never have to ask anyone for verification.
7) If you are trying to get rid of your teacher, he or she may have a heart attack if you submit such a presentation.

In any case, good work on geting the correct answer. Let's do another one! It's pretty amazing what one can learn from a single problem.