Math help aka I'm an idiot

[fire hazard]

So I'm trying to get this math project done and I'm stumped. It's not hard at all I just can't seem to figure it out... Could be because its 5:30am or I'm just missing something(like a brain). I attached a screen shot of what I'm working with. I'm trying to find the formula for getting the area of a rectangle with some given dimensions. I thought I figured it out, but my area comes out as a negative which doesn't work. If someone can help and tell me what I'm doing wrong I will be so fucking grateful. <3

http://imageshack.us/f/545/mathtard.jpg/

Might be useful to put the situation up too...

A construction company wishes to build a rectangular enclosure to store machinery and equipment. The site selected borders on a river that will be used as one of the sides of the rectangle. Fencing will be needed to form the other three sides. The company has 528 feet of 10-foot high chain link fencing. The questions that follow should help you determine the dimensions of the rectangle that will produce that maximum area for the storage site. Assume that the company wishes to use all of the fencing, and that "width" refers to the measure of each of the two sides that are perpendicular to the river, and "length" refers to the measure of the side parallel to the river (see diagram).
The diagram has a rectangle next to a river. The length side closest to the river has a dashed line instead of a solid one. Don't know what the hell thats supposed to mean, but it probably matters.

 

[biostud]

I would say A=-2w^2+528w

 

[DrPizza]

Don't know what the hell thats supposed to mean, but it probably matters.

Did you try reading what you copied and pasted?? It's spelled right out in the text, and yes, it matters.

What have you tried so far? Your equation is close - very close. How did you get that equation? I'm quite surprised you could get that far, yet your post conveys that you're clueless.

 

[fire hazard]

Did you try reading what you copied and pasted?? It's spelled right out in the text, and yes, it matters.

What have you tried so far? Your equation is close - very close. How did you get that equation? I'm quite surprised you could get that far, yet your post conveys that you're clueless.
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I had to read and learn the entire chapter this is on last night before I started. So I sort of know what i'm doing, but it's still so knew that I'm clueless at the same time.

I got that equation this way:

Area of a rectangle is found with: A= L x W
With this data, length is found with: L= 2w-528
Input that into area function: A= (2w-528)w
Distribute: A= 2w^2-528w

 

[Mr. Pedantic]

So what you've written is a quadratic equation. If you try to visualise the equation plotted on a graph, what does it look like?

If you're looking to find the maximum area within the rectangle, what point does this correspond to on the graph? Is there anything special about that point from which you could, say, derive the correct answer?

I had to read and learn the entire chapter this is on last night before I started. So I sort of know what i'm doing, but it's still so knew that I'm clueless at the same time.

I got that equation this way:

Area of a rectangle is found with: A= L x W
With this data, length is found with: L= 2w-528
Input that into area function: A= (2w-528)w
Distribute: A= 2w^2-528w

This isn't quite right.

 

[arosusi]

I had to read and learn the entire chapter this is on last night before I started. So I sort of know what i'm doing, but it's still so knew that I'm clueless at the same time.

I got that equation this way:

Area of a rectangle is found with: A= L x W
With this data, length is found with: L= 2w-528
Input that into area function: A= (2w-528)w
Distribute: A= 2w^2-528w

I think that there's your error. The signs are wrong.

 

[fire hazard]

So what you've written is a quadratic equation. If you try to visualise the equation plotted on a graph, what does it look like?

If you're looking to find the maximum area within the rectangle, what point does this correspond to on the graph? Is there anything special about that point from which you could, say, derive the correct answer?



This isn't quite right.

It should be a parabola that opens downward. The vertex will be the point with the most area. I just can't get that far since I can't even figure out the equation that needs to be graphed.

 

[arosusi]

It should be a parabola that opens downward. The vertex will be the point with the most area. I just can't get that far since I can't even figure out the equation that needs to be graphed.

Let the sides of the rectangle be a and b, b being the one running along the river. Then 2a+b=528. This gives b=528-2a. The area is then A= ab
= a(528-2a) = 528a-2a^2.

 

[skimple]

f(w) = C - 2w, where C is empirically derived as being 528.

w * f(w) = A
w(C - 2w) = A
Cw - 2w^2 = A

-2w^2 +528w = A

 

[fire hazard]

Let the sides of the rectangle be a and b, b being the one running along the river. Then 2a+b=528. This gives b=528-2a. The area is then A= ab
= a(528-2a) = 528a-2a^2.

Yeah, after your first post I messed with the equation and got:
A= -2w^2 + 528w

Thanks for the help. I knew it was something simple, my mind is just jelly right now. I feel like I tried this last night, but I must have input it wrong in the calculator and then wrote it off. Thanks!!!!!!!!!

 

[JTsyo]

heh I remember helping my cousin with this exact question last year.

Area = L*W
Perimeter = 2L+2W ( in this case since you have the river it's just L+2W)

Area = (P-2W)*W Find the max Area (plotting is probably the easiest)

 

[darkewaffle]

Hooray I can still do some math, had to stare at it for a sec but I got it.

 

[TuxDave]

Yeah, after your first post I messed with the equation and got:
A= -2w^2 + 528w

Thanks for the help. I knew it was something simple, my mind is just jelly right now. I feel like I tried this last night, but I must have input it wrong in the calculator and then wrote it off. Thanks!!!!!!!!!

Stop procrastinating and sleep more. Jelly for brains isn't good for learning.

 

[DrPizza]

It's almost ironic about calculus that setting up problems is where 90% of the difficulty lies. The "calculus part" of the problems are generally quite simple.

 

[SheHateMe]

Rule of thumb:

Don't do someone else's homework.

Especially when it is evident from the OP that it's something they should know how to do.

All the OP needs to know is the formulas for rectangles that the problem calls for. Plug in the numbers that are given and use those numbers to find derivatives.

 

[WHAMPOM]

Really,I got A=(528"/3)squared, sure you are not overthinking the problem?

 

[Mr. Pedantic]

Really,I got A=(528"/3)squared, sure you are not overthinking the problem?

You can get more than this.

 

[JTsyo]

Rule of thumb:

Don't do someone else's homework.

Especially when it is evident from the OP that it's something they should know how to do.

All the OP needs to know is the formulas for rectangles that the problem calls for. Plug in the numbers that are given and use those numbers to find derivatives.

OP was almost there. I see no problem in helping when he has already put effort into it and just made a simple mistake.

 

[PlasmaBomb]

Axis of symmetry is x = -b/2a


DrP is there a better way to derive the answer?

 

[SheHateMe]

Axis of symmetry is x = -b/2a


DrP is there a better way to derive the answer?

You're thinking too much about this problem.

All you need is the forumla for Area and Perimeter, a derivative and some plugging in. Then you have it!

 

[DrPizza]

You're thinking too much about this problem.

All you need is the forumla for Area and Perimeter, a derivative and some plugging in. Then you have it!

Psssst. If he uses the equation for the axis of symmetry, it's the same thing. If you take the derivative of ax^2+bx+c, and set it equal to zero, you get x = -b/2a. So actually, he did "less" work by using a formula that 9th grade algebra students should know, rather than doing any calculus.

 

[SheHateMe]

Psssst. If he uses the equation for the axis of symmetry, it's the same thing. If you take the derivative of ax^2+bx+c, and set it equal to zero, you get x = -b/2a. So actually, he did "less" work by using a formula that 9th grade algebra students should know, rather than doing any calculus.

But Dr.Pizza, don't you know that math teachers...especially in College are really hard up about students using methods to solve problems that are not taught in class or in the book?

I took Calc I and II and if I used a simpler way of finding derivatives, I would get ZERO credit. Math teachers are like that, they get all pissy when you don't do things the way they want you to do it.

The point of his assignment is that the problem wants him to use the formula for Area and Perimeter to find the Maximum area that can be covered.

The problem isnt about finding a shortcut, its about knowing how to do a world problem as they have it setup and coming to the conclusion the way they wanted you to.


...math sucks.

 

[Mr. Pedantic]

Psssst. If he uses the equation for the axis of symmetry, it's the same thing. If you take the derivative of ax^2+bx+c, and set it equal to zero, you get x = -b/2a. So actually, he did "less" work by using a formula that 9th grade algebra students should know, rather than doing any calculus.

What's this axis of symmetry thing?

 

[jagec]

What's this axis of symmetry thing?

If you plot area vs. width, you get a parabola. The max area will be at the axis of symmetry, and so you can use the equation for that and save yourself the trouble of plotting.

 

[Mr. Pedantic]

If you plot area vs. width, you get a parabola. The max area will be at the axis of symmetry, and so you can use the equation for that and save yourself the trouble of plotting.

Ah, that's interesting. I never got taught this...