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RE: Dividing by variable when solving algebraic equation

26th October 2016, 11:12

(26th October 2016, 02:48)Alex K Wrote: I'm teaching my 11th graders quadratic functions right now, they are not amused. Question: you have 100 meters of fence to fence in a rectangular area, but one of the sides doesn't need a fence because there's a wall already. So the question is, how long and wide do you choose your corral to maximize the area.

How most physics questions used to sound to me:

A ball is falling from 25m. The mass of the ball is 2kg. Calculate the exact time of the plane landing in Bangladesh if the sun is seen on the north side of the sky and the course of Euro is 2$.

RE: Dividing by variable when solving algebraic equation

26th October 2016, 11:34

I'd estimate 46?

(16th February 2017, 18:16)TheOther JoeFish Wrote: So what you're saying is that I can harass all of the members I want for the next 168 hours, as long as I do so in my signature?

RE: Dividing by variable when solving algebraic equation

26th October 2016, 12:44

(26th October 2016, 07:37)robvalue Wrote: I like questions like that though, because they show a simple application of maths to a real situation where guesswork would only get you so far.

For those that are interested, this is how you get the answer:

You have 100m of fence to make a rectangle against a wall, which forms one of the four sides. So if the length of the rectangle coming away from the wall is x, you'll be using x also for the side above it, leaving 100 - 2x for the third side, opposite the wall.

So the area is

A = x(100 - 2x) = 100x - 2x^2

To find the maximum, we differentiate with respect to x, to find the rate of change in the area. To differentiate Ax^n, you get Anx^(n-1)

dA/dx = 100 - 4x

At a maximum or minimum amount, the rate of change is zero. (Think of the top or bottom of a quadratic curve; the tangent becomes flat.)

0 = 100 - 4x

x = 25

To check this is indeed a maximum, we take the second derivative:

d2A/dx^2 = -4

This means it is a maximum, as its less than zero. The rate of change is lessening as you approach and go over the point. Less than zero is a minimum.

So with x = 25, the side opposite the wall is 100 - 2*25 = 50

Once you get to the formula for A in terms of x, there's an alternate method that doesn't involve using calculus. You can complete the square instead, and then pick the relevant value for x to give what is obviously the maximum value for A.

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RE: Dividing by variable when solving algebraic equation

26th October 2016, 15:47

(25th October 2016, 21:58)FallentoReason Wrote:

(25th October 2016, 21:54)Fireball Wrote: Could I have an example?

Any equation where the constant is positive will do.

E.g. x^2 + 4

Let's test where it intersects the x axis:

x^2 + 4 = 0

=> x^2 = -4

therefore, x = 2i

Thanks! Next time, I'll pour more coffee down the inside of my neck before I ask a question. I taught algebra for a couple of years. My brain obviously had checked out for parts unknown while reading this thread.

RE: Dividing by variable when solving algebraic equation

26th October 2016, 16:01

(26th October 2016, 11:12)Vic Wrote:

(26th October 2016, 02:48)Alex K Wrote: I'm teaching my 11th graders quadratic functions right now, they are not amused. Question: you have 100 meters of fence to fence in a rectangular area, but one of the sides doesn't need a fence because there's a wall already. So the question is, how long and wide do you choose your corral to maximize the area.

How most physics questions used to sound to me:

A ball is falling from 25m. The mass of the ball is 2kg. Calculate the exact time of the plane landing in Bangladesh if the sun is seen on the north side of the sky and the course of Euro is 2$.

RE: Dividing by variable when solving algebraic equation

27th October 2016, 01:32

irrational Wrote:I know for all you math masters out there, this is basic stuff.

I'm far from a being a math master myself. For me at least, I think math is cool and am interested in gaining a basic understanding of it. That being said, I thought your post was pretty cool. I've often found that when I'm struggling with a problem, it is due to my lack of familiarity with the basics, such as having an incomplete understanding of particular terms in definitions or being a bit rusty in fundamental algebra, trig, or calc concepts (etc.). Furthermore, I've learned to value the instances where I have the humility to crack open the books, engage in research, or pick another person's brain: this helps me avoid the trap of using the same problematic thinking, which got me confused in the first place.

P.S. Thanks for the cool thread and good luck with your math endeavors, sir.

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