So here's what I have

 

 

I get lost after the second step, I believe he said to make a perfect square, and then factor into a(x-h)^2+k

Right, you have the more difficult part already. So now we have...

 

A = -2w^2 + 40w

 

Now notice that in a parabola, the maximum (for a downward facing parabola) or the minimum (upward facing parabola) has the x-coordinate as the axis of symmetry. The axis of symmetry can be found by using the following formula:

 

-b/2a

 

The a and b come from the standard form of the quadratic expression

 

Y-coordinate = ax^2 + bx + c

 

In the first equation, the a is -2 and the b is 40. Thusly, the axis of symmetry is -40/2(-2) = 10. Therefore, the x coordinate of the maximum is 10. We can use this to find the y coordinate (which is what you need to solve the problem) by plugging in the x coordinate into all instances of x (which is w, in this case).

 

A = -2(10)^2 + 40 (10)

A = -200 + 400

A = 200

 

I'm unsure if you learned these things or whether I should go ahead with your teacher's way, but your teacher has probaly made this much harder than it needs to be.

I can't find way to get the exact numbers (they don't exist), but we can get as close as:

W = 0,00001

L = 39,99998

A = 0,0004

That may look ok, but it's because at that point excel gives up on exact decimals, as you can see, A is slightly off.

I can't find way to get the exact numbers (they don't exist), but we can get as close as:

W = 0,00001

L = 39,99998

A = 0,0004

That may look ok, but it's because at that point excel gives up on exact decimals, as you can see, A is slightly off.

 

Firstly, it's totally inaccurate. Secondly, Spork needs an algebraic way of solving it for his class. Thirdly, there are exact numbers for them.

Right, you have the more difficult part already. So now we have...

 

A = -2w^2 + 40w

 

Now notice that in a parabola, the maximum (for a downward facing parabola) or the minimum (upward facing parabola) has the x-coordinate as the axis of symmetry. The axis of symmetry can be found by using the following formula:

 

-b/2a

 

The a and b come from the standard form of the quadratic expression

 

Y-coordinate = ax^2 + bx + c

 

In the first equation, the a is -2 and the b is 40. Thusly, the axis of symmetry is -40/2(-2) = 10. Therefore, the x coordinate of the maximum is 10. We can use this to find the y coordinate (which is what you need to solve the problem) by plugging in the x coordinate into all instances of x (which is w, in this case).

 

A = -2(10)^2 + 40 (10)

A = -200 + 400

A = 200

 

I'm unsure if you learned these things or whether I should go ahead with your teacher's way, but your teacher has probaly made this much harder than it needs to be.

Oh, the good ol' -b/2a trick. Forgot about that.

 

IT'S EASIER WITH CALCULUS

NO IT ISN'T

 

INTEGRATION AND CURVE SKETCHING DOESN'T SOLVE EVERYTHING

If you go through my numbers then you'll se that they go perfectly along with the rules stated in the picture. I used the rules for A according to W and used smaller and smaller numbers for W until the A values according to those rules matched.

 

W= 0,00001

A= (40+(A8*2))*A8

A= (A8*40)-((2*A8)^2)

 

P= 40

W= 0,00001

L= F5-(2*F6)

A= F6*F7

 

My excel sheet with the equals removed.

Errr

 

Graphs are considerably easier...

NO IT ISN'T

 

INTEGRATION AND CURVE SKETCHING DOESN'T SOLVE EVERYTHING

Actually, this one's derivative calculus.

 

Maxes and mins have a derivative of zero.

 

So:

A=-2x^2+40x

A'=-4x+40=0

-4x=-40

x=10

 

And I only wrote all that out to "show my work". Normally, I'd just look at it and do the derivative and algebra in my head.

1. Graphs are somewhat inaccurate, but the whole point of the max/min was figuring out the graph.

 

2. Stork, you basically did what I did, but cheated because calculus isn't allowed in geometry class.

So, I might be choosing a signiture soon. What do you guys think?

Kewl.

Any suggestions though :P.

2. Stork, you basically did what I did, but cheated because calculus isn't allowed in geometry class.

But... but... I LIKE calculus...

 

Now if it were a trig problem, I'd say "screw it, you're on your own." Trig is evil.

Trig is great. Messing around with the identities for a few minutes while the clock is ticking on the tests is the definition of a rollercoaster ride.

Trig sucks, sadly it is used in several aspects of electrics, so I have to use it a lot. But luckily it is only used in multi-fase currents and I'll be rid of it next year in electronics.

I hate trigonometry, which is part of the reason I am not so great at vectors.

I hate trigonometry, which is part of the reason I am not so great at vectors.

Vector trig is easier than calculus trig. I mean, it's just sine and cosine relative to your coordinate plane, as far as I've learned.

 

Calculus trig gets annoying because you have to chain-rule the trig functions. And if you're already doing the chain rule once or twice, it gets really easy to screw up.

Sin, Cos, Tan, Sin^-1, Cos^-1 and Tan^-1 to calculate the sides and agles of a 90° triangle according to eachother. It's hell no matter how you use it. And you will only need it in architecture and AC motors, yet we still have to learn it.

Well I love trigonometry, its just a set of rules that you follow. Nothing too hard, no actual brain work.

Mathmatics is basically following a set of rules to a desired outcome...hence why computers can do it far better than humans.

 

The only area Humans excel over computers is deciding which set of rules to use.

[bleep] math.

No there's no brain work, because it is impossible without a calculator or computer, and I mean literally impossible.

[bleep] math.

 

This. A million [bleep]ing times.

 

Seriously, [bleep] math.

 

 

And I'm having to take remedial algebra because my high school teachers sucked (except my Algebra I teacher, but I took his class in 8th grade and forgot a lot of what I learned). :wall:

[bleep] math.

 

This. A million [bleep]ing times.

 

Seriously, [bleep] math.

Indeed.