
so I'm having trouble with optimization I don't think I understand it well enough. I know the whole two formula plugin and derive, but I don't think I'm getting it.
You have 20 ft of wire, and you need to make an equilateral triangle and a square. How much should be used for each figure so that the area is maximized?
x for square, b for triangle so far, I have: 4x+3b=20 x^2+1/2(bh)=A h=root(b^2[b/2]^2)
x=53/4(b)
=> 2530/4(b)+9/16(b^2)+b/2(root[b^2(b/2)^2])
it seems like it's way too complicated.. Am I doing it wrong?
might be more questions later ;;
ok, problem 2:
Pink Pig is in the muddy field 4 mi from the nearest point, P, on Farm Road 5. She wants to get to Sally's Slop House on that road, 7 mi to the right of point P in the shortest possible time. Since her maximum possible speeds are 3 mph in mud and 6 mph on the road, she mentally calculates that she must maintain maximum possible speed and intersect the road ___ mi to the right of point P.


oh god, i sucked at calculus, and i hated this section the most lol.

Find the derivative of the formulas and solve for X when f(x) = 0

yeah, but taking the derivative of the second part is going to be a mess t_t and the book usually points out an easier way. Just wanted to see if there is a smarter way, or if I just have to brute force this shit.

That's the most they can throw at you for Calc 1, annoying derivatives, suck it up and doing it is probably the only way.
Though that kind of derivative looks pretty normal for something from optimization.

auuuuuugghhhh i hate algebra zzzzz
thanks for the help though : )

if you're in a hurry you can always use those online derivative calculators... but i'd recommend you use those as a ways to check your work, for I assume your teacher won't allow you to use the internet while taking the test

United States23604 Posts
I just slapped something together and the calc was trivial compared to the other stuff:
For starters I assumed the length of the string is L, the perimeter of the square is p1, the perimeter of the triangle is p2, and p1+p2=L.
Area1 = (p1 / 4)^2 = p1 / 16 Area2 = .5 * (p2 / 3) * h where I used the law of sines to get h = p2*sqrt(3)/6
So A = A1+A2 Take the derivative of the quadratic (easy if you set i up decently) to get a linear equation, and then set it equal to zero to see if there is a max/min/etc...
And I get that the triangle should be 20/(1+sqrt(3)/2) and thus the square should be 20(11/(1+sqrt(3)/2))

Kentor
United States5784 Posts
Let's say
s = side of square and b = side of triangle
4s + 3b = 20, or b = (20  4s)/3, or s = (20  3b)/4
For an equilateral triangle, the height, in terms of its base, is h = root(3)/2 * b So the area of the triangle is 1/2 * b * h = root(3)/4 * b^2
Total area will be
s^2 + root(3)/4 * b^2 = A
substitute either s in terms of b, or b in terms of s, let's just choose the first one
((20  3b)/4)^2 + root(3)/4 * b^2 = A (9b^2  120b + 400)/16 + root(3)/4 * b^2 = A 9/16 * b^2  15/2 b + 25 + root(3)/4 * b^2 = A
differentiate and solve for b
(9/8 + root(3)/2)b  15/2 = 0
the numbers aren't gonna be pretty. just need to do it.
I think the only thing that's making it seem more complicated than it is is the way you expressed the area of an equilateral triangle.


ahhh I see. Thanks! I must have missed geometry when the teacher was talking about it :o
and Micronesia.. i didn't understand your post : (

On January 21 2009 11:32 xMiragex wrote:this + Show Spoiler +http://img.4chan.org/b/src/1232496239730.jpg Lol linking a 4chan image is the stupidest thing you can do since they go 404 so quickly.

I fucking hate this shit; I think I just bombed a test on this :/

On January 21 2009 10:32 mahnini wrote: 1a2a3a HAHAHAHAHAHHAA
ROFL
On January 21 2009 11:33 redtooth wrote:Lol linking a 4chan image is the stupidest thing you can do since they go 404 so quickly.
i can think of a lot of things more stupid than that

United States23604 Posts
On January 21 2009 11:32 FragKrag wrote: ahhh I see. Thanks! I must have missed geometry when the teacher was talking about it :o
and Micronesia.. i didn't understand your post : ( That's ok... I wasn't that clear since I did it quickly. I'm not sure if my method of using the perimeter is better or not anyway.

ok, problem 2:
Pink Pig is in the muddy field 4 mi from the nearest point, P, on Farm Road 5. She wants to get to Sally's Slop House on that road, 7 mi to the right of point P in the shortest possible time. Since her maximum possible speeds are 3 mph in mud and 6 mph on the road, she mentally calculates that she must maintain maximum possible speed and intersect the road ___ mi to the right of point P.
It's a mental problem so no scratch paper or calc. T_T I think it incorporates related rates :x
updated OP

Well, taking an intuitive approach to this problem is better than solving an ugly equation. Consider a unit equilateral triangle. That is, each side = 1. The area by well known formulas is going to be the square root of 3 divided by 4 or about .433. The square of perimeter 3 is going to have each side equal to 3/4. So the area is 9/16. 9/16 > .433. Now we know that every square and equilateral triangle is similar to these so the square is always going to be a more efficient use of wire by that ratio if you're looking for area. So you should use it all on the square.

Question 2: sqrt(3) miles. (see edit)
MS paint awesomeness:
EDIT: oh, fuck, I put the pig 3 miles from the road instead of 4... the same idea works, but change sqrt(9 + x^2) to sqrt (16 + x^2) and rework it. x should end up being, uhh... sqrt(16/3).

A 4 miles, what's this mile you speak of?
Anyway, good luck with Optimization, my worst section of Calc in high school.

On January 21 2009 13:24 ninjafetus wrote:Question 2: sqrt(3) miles. (see edit) MS paint awesomeness: EDIT: oh, fuck, I put the pig 3 miles from the road instead of 4... the same idea works, but change sqrt(9 + x^2) to sqrt (16 + x^2) and rework it. x should end up being, uhh... sqrt(16/3).
Ah, ok. It makes sense to me now. Thanks~

hmmm... for the 2nd question I got an answer I wasn't expecting but the math kinda got big so either I messed up or it is a "messy" problem.
I got the answer to be 2.3094 miles to the right of point P.
This is what I did.
1. Name distance from P toward destination S (for Sallys) wheverever you intersect the road as L. This is what we are trying to solve for.
2. Express the distance traveled on the road or through the field in terms of L so you only have 1 variable. Lets have distance in mud be y and distance on road be z.
3. z = 7L and x=(L^2 + 16)^.5 (remember that .5 is just the square root)
4. Write equation for time so you can optimize it. That is just z/3 + x/6 since time = distance/velocity.
5. Take derivative and set to 0. Don't forget chain rule on x. It gets messy here.
6. Solve for L. I got L to be 2.3094. or the square root of 16/3

I think there is an error in the paint version above my post. It looks like they set distance to P as 3. Thats how they got 9 while in my equation I got 16.
EDIT: Ahh crap. I thought I was awesome but I just noticed the edit. So I wasn't first. But at least you have 2 independent answers both coming to sqrt or 16/3 so you can be confident.

i think i am understanding it more now : )
<3 TL <3

PM me if you ever want calc/optimization help but don't wanna make a new thread. I enjoy helping with this kind of stuffI used to tutor for free.



