Writing Math Paper 1 on Friday - Help Please

**janvdl** · August 27th 2007, 07:04 AM

Hey Guys, i write math paper 1(Algebra) on friday and i'll post questions in here when i need help.

Okay the first one i need help with (

) :

I attached a rough sketch with the prob.

All that we are given is the equation of the pink line (y = -2x + 18) and that one right angle.

The question asks to find the maximum area of the red triangle.

This one really bowled me over, please help guys.

**TKHunny** · August 27th 2007, 08:41 AM

Call the point where the triangle meets the line (a,b).

The area of the triangle is ½*a*b, right?

Because (a,b) is on the line, we are given that b = -2a + 18.

Substitute.

Area(a) = ½*a*(-2a + 18)

Now, what do you know about parabolas with negative leading coefficients?

**janvdl** · August 27th 2007, 08:51 AM

Oh wait so we can say that since $b = -2a + 18$ and $A = 0,5(a)(b)$ that:

$A = (0,5)(a)(-2a + 18)$ ?

$A = (0,5)(-2a^2 + 18a)$
$A = -a^2 + 9a$

$\frac{dA}{da} A = -2a + 9$
$0 = -2a + 9$
$a = 4,5$

**TKHunny** · August 27th 2007, 08:58 AM

Well, if you insist on breaking out the calculus, but that's a little much for this problem. Your analytic geometry should get you through this one. On the other hand, if you are in a calculus section, I guess that would be the way to proceed.

One word of caution, don't forget your Domain. If your result says a = 10, then you need to keep thinking about it.

**janvdl** · August 27th 2007, 09:02 AM

Originally Posted by TKHunny

Well, if you insist on breaking out the calculus, but that's a little much for this problem. Your analytic geometry should get you through this one. On the other hand, if you are in a calculus section, I guess that would be the way to proceed.

One word of caution, don't forget your Domain. If your result says a = 10, then you need to keep thinking about it.

I am supposed to use Calculus on this...

**janvdl** · August 27th 2007, 09:11 AM

Ah, now this problem is easy.

Thank you so much for your help.

**ThePerfectHacker** · August 27th 2007, 09:29 AM

Originally Posted by janvdl

I am supposed to use Calculus on this...

They do Calculus in 11th grade (or 10th grade) in South Africa?

Originally Posted by janvdl

Ah, now this problem is easy.

What TKHunny did is the general idea. Let "x" represent the quantity and find what "x" gives the maximum.

You want a much more difficult question? It is really fun.

(This thread belongs in Calculus not in High School).

**janvdl** · August 27th 2007, 09:34 AM

Originally Posted by ThePerfectHacker

They do Calculus in 11th grade (or 10th grade) in South Africa?

I'm in Grade 12.

My Final year!!!!!!!! Yes!!!!!

Originally Posted by ThePerfectHacker

You want a much more difficult question? It is really fun.

As long as you don't kill me with it

Okay, i'm game, hit me with your best shot.

Originally Posted by ThePerfectHacker

(This thread belongs in Calculus not in High School).

But i am in high school...

**ThePerfectHacker** · August 27th 2007, 09:54 AM

Originally Posted by janvdl

I'm in Grade 12.

My Final year!!!!!!!! Yes!!!!!

Why are you so happy about it. I never wanted to graduate from high school.

As long as you don't kill me with it.

Let a wire have length $L$ . You cut the wire. You take one part of it and turn it into a perfect circle. You take the other part of it and turn it into a perfect square. Where should you cut the wire to minimize the total area obtained. And where should you want to cut the wire to maximize the total area obtained. (Spend time on this problem).

**janvdl** · August 27th 2007, 10:05 AM

Originally Posted by ThePerfectHacker

Why are you so happy about it. I never wanted to graduate from high school.

Because i hate school.

Let a wire have length $L$ . You cut the wire. You take one part of it and turn it into a perfect circle. You take the other part of it and turn it into a perfect square. Where should you cut the wire to minimize the total area obtained. And where should you want to cut the wire to maximize the total area obtained. (Spend time on this problem).

Okay we can say that:

$Circumference \ of \ Circle = 2 \pi r$

$L = 2 \pi r$

$r = \frac{L}{2 \pi}$

$Area \ of \ Circle = \pi ( \frac{L}{2 \pi} )^2$

----

$Area \ of \ Square = L^2$

Please tell me if i'm on the right path here.
I'm thinking of multiplying or maybe adding these two. Is that correct?

----

$\pi ( \frac{L}{2 \pi} )^2 + L^2 = \frac{L^2}{4 \pi} + L^2 = \frac{L^2 + 4 \pi (L^2)}{4 \pi}$

But that must equal 2A ?

So: $A = \frac{L^2 + 4 \pi (L^2)}{8 \pi}$

I think we can differentiate now...

I get $\frac{dA}{dL} = \frac{\pi}{4} L + L$

Set equal to 0

$0 = \frac{\pi}{4} L + L$

But then L = 0...

**janvdl** · August 27th 2007, 10:20 AM

I guess that's way off eh TPH?

**ThePerfectHacker** · August 27th 2007, 10:40 AM

Hint: Let $x$ be the distance away from the left end up the wire where you cut. Express the area function in terms of $x$ .

**janvdl** · August 27th 2007, 11:39 PM

OK, let's cut $L$ into $x$ and $L - x$

$Area_{Circle} = \frac{x^2}{4 \pi}$

$Area_{Square} = L^2 - 2xL + x^2$

$Area_{Circle+Square} = 4 \pi x^2 + 2 \pi L^2 - 4 \pi x L$

But L is a constant, so now we can easily differentiate.

$\frac{dA}{dx} A = 8 \pi x - 4 \pi L = 0$

$x = \frac{L}{2}$

**DivideBy0** · August 27th 2007, 11:43 PM

Are you sure $Area_{Circle+Square}$ is right?

**janvdl** · August 27th 2007, 11:47 PM

OK, let's cut $L$ into $x$ and $L - x$

$Area_{Circle} = \frac{x^2}{4 \pi}$
$Area_{Square} = L^2 - 2xL + x^2$

$Area_{Circle+Square} = 2A$

$A = 4 \pi x^2 + 2 \pi L^2 - 4 \pi x L$

But L is a constant, so now we can easily differentiate.

$\frac{dA}{dx} A = 8 \pi x - 4 \pi L = 0$

$x = \frac{L}{2}$

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