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RE: Dividing by variable when solving algebraic equation

26th October 2016, 02:32

Keep up the good work everyone

"Leave it to me to find a way to be,
Consider me a satellite forever orbiting,
I knew the rules but the rules did not know me, guaranteed." - Eddie Vedder

RE: Dividing by variable when solving algebraic equation

26th October 2016, 02:48 (This post was last modified: 26th October 2016, 02:49 by 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.

The fool hath said in his heart, There is a God. They are corrupt, they have done abominable works, there is none that doeth good.

RE: Dividing by variable when solving algebraic equation

26th October 2016, 03:04

(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.

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.

"It is the mark of an educated mind to be able to entertain a thought without accepting it" ~ Aristotle

RE: Dividing by variable when solving algebraic equation

26th October 2016, 07:37 (This post was last modified: 26th October 2016, 07:38 by 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

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