Calculus Optimization Problem - Need Serious Help

Printable View

12-07-2004, 11:15 PM
Kirobaito

1 Attachment(s)

Calculus Optimization Problem - Need Serious Help

Okay, we're going over optimization in Calculus right now. I understand it perfectly when my teacher does it on the board; I know exactly what he's doing and why. But every time he gives us a problem to work on our own I get lost and end up going in circles and the like. We got one, and I've tried it a hundred times, and can never make it easier. Help would be appreciated. My work so far is shown.

AP CALCULUS
FIND THE DIMENSIONS OF THE LARGEST ISOSCELES TRIANGLE THAT CAN BE INSCRIBED IN A CIRCLE OF RADIUS 4.

OK, so pretty much this is how I've drawn it up.

Okay, Photobucket is being a meanie, so it's now attached.

Now, here are my formulas that I've made. I'm attempting to make the Area formula contain only one variable, in this case x.

A=xy
The formula for the area in this problem, since the base is actually 2x. Makes it easier.

y=4+h
Simple logic.

h^2+x^2=16
h=(16-x^2)^(1/2)
OK, the formula for h in terms of x, so I can replace it in the area formula. I used the pythagorean theorem.

A=x[4+(16-x^2)^(1/2)]
OK, this is the area formula, and it only contains X. Now, before I find the derivative in terms of X so I can find the Maximum of the A(x) curve, I'm going to distribute the x.

A=4x+(x)(16-x^2)^(1/2)
I'm correct so far, yes?

Now, here's where it all gets messy. I'm going to find the derivative in terms of X, and then set dA/dx equal to zero to find the maximum.

A=4x+(x)(16-x^2)^(1/2)
dA/dx=4+....
OK, now here's me using the product and chain rules.
[(1)(16-x^2)^(1/2)+x(1/2)(16-x^2)^(-1/2)(0-2x)]

That simplifies to:

dA/dx=4+[16-x^2)^(1/2)+x(1/2)(16-x^2)^(-1/2)(-2x)]

Now, the 1/2 and the 2 in the 2x at the end can cancel. So make it:

dA/dx=4+[16-x^2)^(1/2)+x(16-x^2)^(-1/2)(-x)]

I'm gonna combine the -x and x now to make -x^2.

dA/dx=4+(16-x^2)^(1/2)+(-x^2)(16-x^2)^(-1/2)

Here's where it gets so messy it begins difficult to type.

0=4+(16-x^2)^(1/2)+[(-x^2)/[(16-x^2)^(1/2)]]

Does that make sense? Now, I'm gonna multiply every value by (16-x^2) so I can get it out of the denominator on the final term. It would help if you had been writing this, if anybody chose to help me.

0=4(16-x^2)+(16-x^2)[(16-x^2)^(1/2)]-x^2

Is this correct AT ALL?

The work gets way to messy and it's here that I start to doubt myself. Have I done any of this right?
12-07-2004, 11:32 PM
Necronopticous

Let me get back to you next semester :(
12-07-2004, 11:33 PM
Kirobaito

Too late, this is due tomorrow. ;D
12-08-2004, 01:30 AM
crono_logical

Since the height of the triangle is what it looks like we're going to vary, why not do the equation in terms of what you've labelled as y? Certainly looks easier that way instead to me, without grabbing pen and paper :p My guess though would be that the answer is an equilateral triangle :p
12-08-2004, 03:19 AM
Kirkpatrick

Yeah, I'd guess an equilateral triangle too - just like in rectangle optimization questions where it's always a square :P

I'm also working on this, I'll get back to you if I figure something out. It's good review since we did optimization a few weeks ago.
12-08-2004, 03:24 AM
Necronopticous

I can't wait for next semester, this looks so interesting.
12-08-2004, 03:51 AM
Kirobaito

My method certainly didn't yield anything.

I came up with -23x^2+x^4=128, which gets you NOTHING.

I suck at Calculus. UNNE, help me.
12-08-2004, 05:57 AM
Del Murder

I have a formula. Assuming b and h are the base and height of the triangle:

b^2 = 32h - 2h^2

Substitute that into the area formula and then use derivatives, blah blah you should be able to do that part.

I do not know if this is correct. I got this by drawing the picture and making different triangles. I noticed something in the triangles I made that looked good but I didn't have time to go check, because it's almost bedtime. I hope this gets you on the right track, though. What I do know is that your first method looks completely wrong. :D

This is as much you get from me by whining for Unne first instead your friendly neighborhood math graduate.
12-08-2004, 06:13 AM
Dr Unne

I would agree that your [EDIT: meaning the first post's] method looks completely wrong. But it's sort of been 4 years since I cracked open any calc book. It's been almost 6 years since I've done an optimization problem. I'm not going to be much help. I'm going to type it as I think it.

Optimization problems involve finding the minimum and maximum of a given function. Easiest way to do that is often to graph it. The other option is to find a place where the derivative is 0 (right?) because when the slope of the tangent line is 0, the graph is at a min or max. That should be the only place you need to use calculus.

Getting the right equation is the hard part. My HS calc teacher told us that you can't learn how to do that. It's creativity, and either you get it or you don't. I didn't buy that.

It should be obvious that a triangle inside a circle with radius 4 can't have a side that's longer than or equal to 8. If it did it wouldn't be a triangle.

This equation is the area of an isosceles triangle with base a, height h, and two equal sides b:

If you can find a way to get b in terms of a, then you have an equation with two variables: area and base. As the base changes, the area changes. Maximize area by letting the base take on all possible values and finding the minimum and maximum of that function. You can almost certainly find b, given a, by the fact that when you draw a chord with length a, if you know the radius of the circle, you can find the length of the two arcs on either end of the chord. Find the length of the bigger arc, and then take half of that length, and then find the chord that is defined by that arc, and there you go, you now have b.

This is so complicated that it probably is not the easiest way to do this, if it's even possible. Maybe it'll give you some ideas. Maybe (more likely) it'll give you some ideas on what not to do. Next time ask the neighborhood math graduate.
12-08-2004, 06:24 AM
Del Murder

1 Attachment(s)

Here is a crude drawing of how I came up with my formula. That is all the help I will give. Good luck!
12-08-2004, 09:26 AM
escobert

Quote:

Originally Posted by Necronopticous

I can't wait for next semester, this looks so interesting.

:eek:

I'm so horrible at math.

What happend to the homework forum?
12-08-2004, 10:54 AM
crono_logical

Not enough people did homework that it needed its own entire forum :p
12-08-2004, 03:53 PM
Endless

Let's call X the angle between c and h (look at Del's picture).

c = 8*cos(X) [cos(X) = c/8]
b/2 = c*sin(X) [sin(X) = (b/2)/c]
b/2 = 8*cos(X)*sin(X)
h = 8*cos(X)*sqrt(1-sin(X)^2)
h = 8*cos(X)^2

So, A = h*b/2 = 64*cos(X)^3*sin(X)
Transformate that horror:
A = 64*cos(X)^2*(0.5*sin(2*X))
A = 64*(1/2)*(1+cos(2*X))*(1/2)*sin(2*X)
A = 16*sin(2*X)+8*sin(4*X)
(find your trig book, there are a lot of ways to manipulate cos and sin, and you do end up with this)

Derivate that (fun xD)
A' = 32*(cos(2*X)+cos(4*X)) = 0
cos(2*X)+cos(4*X) = 0
Another transformation
cos(pi-2*X) = cos(4*X)
pi-2*X = 4*X
X = pi/6

From this you find:
c=4*sqrt(3)
b=4*sqrt(3)
h=6
A=12*sqrt(3)

We now have the mathematical justification of the obvious (to me) result to your question: the biggest isoceles triangle that fits in a circle is an equilateral triangle.

Edit: just reread Del's posts. I think you get:
h = (32-(b/2)^2-(b/2)^4)^(1/4)
A = hb/2
A = (1/2)*b*(32-(b/2)^2-(b/2)^4)^(1/4)
And have fun with solving that derivate = 0.

Edit^2:
Reread the orginal post, whose method is right.
A=x*(4+sqrt(16-x^2))

Derivate that, you get:
4+sqrt(16-x^2)-x*(x/(sqrt(16-x^2))) = 0
4+sqrt(16-x^2) = (x^2)/(sqrt(16-x^2))
x != 0, x != -4, x != 4
4*sqrt(16-x^2) + 16 - x^2 = x^2
sqrt(16-x^2) = 1/2 * x^2 - 4
16-x^2 = (1/2 * x^2 - 4)^2
16 = 1/4 * x^4 - 3* x^2 +16
(1/4) x^4 = 3 * x^2
So... 1/4 x^2 = 3
x^2=12
x=2*sqrt(3)
12-08-2004, 08:40 PM
Kirobaito

Well, I eventually tried it again this morning, and got the answer. My method was completely right, strangely enough, doing what I was taught. I just picked the wrong variable to try and differentiate with. Doing it with what I had labeled as h is much easier than doing it with x or y.

This time I found the Area in terms of what I had labeled as h.

A=xy

Duh.

Well, x^2+h^2=16. Duh.

So x=(16-h^2)^(1/2).

Now I need to get y in terms of h. Obviously, y=h+4.

So my new equation that I used it this.

A=[(16-h^2)^(1/2)][h+4]

Now, find the derivative in terms of h, and then I'll set it to 0 to determine what h will be equal to at the maximum area.

dA/dh=(1/2)[(16-h^2)^(-1/2)](-2h)(h+4)+(1)(16-h^2)^(1/2)

This is really hard to type out. :p

OK, the (1/2) and the (-2h) can be combined, and just combining the first part of the equation gives you:

[(-h^2-4h)]/[(16-h^2)^(1/2)]+(16-h^2)^(1/2)=0

Now, I'm going to multiple everything by (16-h^2)^(1/2) to not only get rid of the radicals, but also get rid of the denominator in the first half of the equation.

[(-h^2-4h)]/[(16-h^2)^(1/2)] multipled by (16-h^2)^(1/2) just gives you (h^2-4h). Easy. (16-h^2)^(1/2) squared is just (16-h^2).

So the new equation is:

-h^2-4h + 16-h^2=0

-2h^2-4h+16=0

Factor out a -2

-2(h^2+2h-8)=0

The -2 can be discarded because it's a constant, and factoring the second part gets you:

(h+4)(h-2)=0

Obviously, h cannot equal -4, so h=2. Yay!

Doing the Pythagorean Theorem you get x^2+2^2=16, blah blah blah, x=2sqrt3. The entire base is 4sqrt3. The height is 6. And since Endless proved that it's an equilateral triangle, the / and \ are also 4sqrt3.

And I did all by myself this morning, too. Unfortunately I went to bed before I read any of you guys' help xD but thanks. :D
12-08-2004, 09:39 PM
SeeDRankLou

Nicely done.
12-09-2004, 01:51 AM
Del Murder

Reading math on a forum is murder. I worked it out at work today and got what everone else got, even though I still couldn't prove why that angle is a right angle. Anyway, it is, but that wasn't necessary for your method, which is good.

Glad you figured it out on your own, though. Good work!
12-09-2004, 01:59 AM
crono_logical

Since when did equilateral triangles have right angles in? :p
12-09-2004, 02:23 AM
escobert

Never. Yay the only math I was good at was geometry, I should have stopped after geometry. Damn PSU making me pass algerbra II before i could be accepted :(
12-09-2004, 03:56 AM
Del Murder

Quote:

Originally Posted by Farah

Since when did equilateral triangles have right angles in? :p

Since never. Maybe you should look at my drawing and see what I referred to before making a smart alec comment, Archie!
12-09-2004, 09:37 AM
Auronhart

Hmm, too bad I missed out, I like math problems. :)
12-09-2004, 08:39 PM
SeeDRankLou

Quote:

Originally Posted by Del Murder

even though I still couldn't prove why that angle is a right angle.

In your picture, we know that the (b/2)(8-h)z triangle is a right triangle. And with the problem solved we know that b/2=2√3 and that 8-h=2, therefore z=4. Also with the problem solve one can also derive that c=4√3. So (d=diameter), if the zcd triangle is a right triangle, then the Pythagorean theorem should work. And behold 4² + (4√3)² = 8². Therefore, zcd is a right triangle. Without having solved the problem I'm sure you can do what Kirobaito did and put everything in terms of h and use the same logic as above.
12-09-2004, 11:46 PM
Endless

Any triangle whose vertex is on the edge of the circle and has the diameter as the base is a right triangle. You can consider it a given.

Graphically, it's piss-easy to demonstrate it. Take your vertex, draw the line that starts from it, crosses the center of the disk, and mark a vertex where it touches the opposite edge. You now have 4 vertices, the original one, the one you just drew, and the two on the diameter. any quadrilatere with its diagonals of equal lengths, cutting each others in their middle is a rectangle, which means, right angles.