Dividing by variable when solving algebraic equation

[Arkilogue]

Keep up the good work everyone

[Alex K]

And here I go with an imaginary heart attack.

Would it kill you if I said I have an engineering degree...? *gulp!*

In my defence, I study architecture at the moment. Maybe maths just wasn't meant to be 

"Hmm I could have sworn that I only planned this house with one front door... I wonder what happened!"

[FallentoReason]

Would it kill you if I said I have an engineering degree...? *gulp!*

In my defence, I study architecture at the moment. Maybe maths just wasn't meant to be 

"Hmm I could have sworn that I only planned this house with one front door... I wonder what happened!"

Wow! Lightbulb moment.

Imaginary doors.


/architect

[Alex K]

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.

[robvalue]

Ooh, ooh, I know!

---

25 by 50

---

[FallentoReason]

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.

Hey! That's not fair. I've done my fair share of maths. You're not allowed to slip me that question 

Then again, I only got 50% in my own example.

[Alex K]

Ooh, ooh, I know!

---

25 by 50

---

YOU WIN THIS TIME, ROBERTO

[robvalue]

If I don't win, the game is rigged.

In case it's not obvious, I love maths Oh yeah. Always have. I asked for homework at my first school.

[robvalue]

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

---

[Longhorn]

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

Also x = -2i

SEE WHAT I MEAN?!

are you OK, Rob o_o