The worlds first riddle!

NERVOUS INSOMINIA Smile

(oh the Brit is on a roll - happy days Very Happy

)

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Izzie:

INQUISITIVE MINDS Cool

INTERESTING DILEMMA Cool

NERVOUS INSOMINIA Cool

You're good!

Ok here are a few more to keep you busy for the next 5 minutes!

TUMOITTHEIROSN Question

CTOHIHNEKRIENNGT Question

DIGAESCTIIODN Question

Now this nervous insomniac needs to try to get some sleep. See you tomorrow. Very Happy

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Izzie wrote:

NERVOUS INSOMINIA

(oh the Brit is on a roll - happy days )

Aren't you any good! Wow, how clever.

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MOTHERS INTUITION Smile

INCOHERENT THINKING Smile

ACID INDIGESTION Crying or Very sad

(needs PeptoBismol Twisted Evil

)

(don't worry Stormy - just keep doing what you do, you make a difference Smile

)

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raprap wrote:

[size=7]Complete the triangles
the 10 ft ladder is 6:8:10
the 12 ft ladder is 6:6*3^.5:12
at intersection the triangles create to similar triangles to ladders
the height of both triangles is d, the bases are r and (6-r)
since triangles are similar to ladders on is
r/d=6/(6*3^.5)
rearranging
r=d/3^.5
the other is
(6-r)/d=8/6
r=(18-4d)/3
equate
d/3^.3=(18-4d)/3
solve for d
d=18*3^.5/(3+4*3^.5)
this is messy
working through some tedious arithmetic
d=(4+3^.5)/5 ft[/size]

Rap

Arithmetic problems

height is 24(9-4*3^.5)/11 = 4.52 ft

Rap

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Rap:

Ladders
height is 24(9-4*3^.5)/11 = 4.52 ft Cool

And so it would appear to be

Consider if you will the two ladders as two lines on a graph. Let the shorter ladder extend from (0,0) to (6,8). Let the longer ladder extend from (6,0) to (0,108^1/2). The square root of 108 can be found using the pythagorean formula. Then solve for the slope and y intercept to find the equations of the two lines:

The shorter ladder have the equation y=(4/3)*x.
The longer ladder has the equation y=(-108^1/2/ 6)*x + 108^1/2.
The lines meet where the ladders cross.
Use substitution to solve for y.
The answer is 108^1/2/(1+(108^1/2/ 8)) =~ 4.52 feet.

If x and y are the heights where the ladders touch the wall, and p is the height of the intersection then 1/p = 1/x + 1/y.

To prove this let x be the height on the right wall and y the height on the left wall. Call a the distance from the left edge of the alley to the point on the alley directly below the intersection point of the ladders. Call b the distance from the right edge of the alley to the point on the alley directly below the intersection point of the ladders.

From similar triangles we get:
p/x = a/(a+b)
p/y = b/(a+b)
Add the two equations:
p/x + p/y = a/(a+b) + b/(a+b)
py/xy + px/xy = (a+b)/(a+b)
p(x+y)/xy = 1
p = xy/x+y
1/p = x+y/xy
1/p = x/xy + y/xy
1/p = 1/y + 1/x

Better yet; carry a tape in your pocket and measure on site! Laughing

tedarsnecgetriouosn Question

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tedarsnecgetriouosn = dangerous intersection

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Stormy:

tedarsnecgetriouosn = dangerous intersection Cool

Way to go! Laughing

I have a problem (no, not that one!); my factory produces tables and chairs is equipped with 10 saws, 6 lathes, and 18 sanding machines.

It takes a chair 10 minutes on a saw, 5 minutes on a lathe, and 5 minutes of sanding to be completed.

It takes a table 5 minutes on a saw, 5 minutes on a lathe, and 20 minutes of sanding to be completed.

A chair sells for $10 and a table sells for $20.

How many tables and chairs should the factory produce per hour to yield the highest revenue Question

, and what is that revenue Question

Any help appreciated!

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Tables and Chairs
[size=7]Total Machine Time available
Saws-600 min
Lathes-360 min
Sanders-1080 min
Let C=# chairs, T=# Tables, P=$ Profit
P=10C+20T
Differentiate & maximize (dP=0)
dP=10dC+20dT=0
so
dC/dT=-2
Integrate
C=-2T+K(onst)

Setting Time Constraints
Saws
10C+5T<=600
10(-2T+K)+5T<=600
10K-15T<=600
2K-3T<=120
2K<=120+3T

Lathe
5C+5T<+360
5(-2T+K)+5T<=360
5K-5T<=360
K-T<=72
2K<=144+2T

Sander
5C+20T<=1080
5(-2T+K)+20T<+1080
5K+10T<=1080
K+2T<=216
2K<=432-4T

2K<=120+3T
2K<=144+2T
2K<+432-4T

Subtract 1 from 2
0<=24-T
T<=24

Subtract 1 from 3
0<=312-7T
T<=44 4/7
Subtract 2 from 3
0<=288-6T
T<=48

Three Possibilities
24 tables per hour
44 Tables per hour
48 Tables per hour

Working through Machine Availability this is
24 Tables 48 Chairs P=$960/hr
44 Tables 28 Chairs P=$1080/hr
48 Tables 24 chairs P=$1200/hr

So the most profitable is 48 Tables and 24 Chairs per hour for $1200 per hour[/size]

Rap
.

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TABLES & CHAIRS

[size=7]Maximize 10C + 20T subject to:
2C + T <= 120
C + T <= 72
C + 4T <216>= 0
T >= 0

The solution space is a pentagon bounded by the lines described above.
The revenue function will be maximized at a vertex of the solution space.
The vertices (intersections of the boundary lines) and their associated revenues are (C,T):
(0,0): $0
(60,0): $600
(48,24): $960
(24,48): $1200
(0,54): $1080

Revenue is maximized at $1200 with 24 chairs and 48 tables.[/size]

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LADDERS

Riding high?

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Rap:

Tables and Chairs
Total Machine Time available
Saws-600 min
Lathes-360 min
Sanders-1080 min
Let C=# chairs, T=# Tables, P=$ Profit
P=10C+20T
Differentiate & maximize (dP=0)
dP=10dC+20dT=0
so
dC/dT=-2
Integrate
C=-2T+K(onst)

Setting Time Constraints
Saws
10C+5T<=600
10(-2T+K)+5T<=600
10K-15T<=600
2K-3T<=120
2K<=120+3T

Lathe
5C+5T<+360
5(-2T+K)+5T<=360
5K-5T<=360
K-T<=72
2K<=144+2T

Sander
5C+20T<=1080
5(-2T+K)+20T<+1080
5K+10T<=1080
K+2T<=216
2K<=432-4T

2K<=120+3T
2K<=144+2T
2K<+432-4T

Subtract 1 from 2
0<=24-T
T<=24

Subtract 1 from 3
0<=312-7T
T<=44 4/7
Subtract 2 from 3
0<=288-6T
T<=48

Three Possibilities
24 tables per hour
44 Tables per hour
48 Tables per hour

Working through Machine Availability this is
24 Tables 48 Chairs P=$960/hr
44 Tables 28 Chairs P=$1080/hr
48 Tables 24 chairs P=$1200/hr

So the most profitable is 48 Tables and 24 Chairs per hour for $1200 per hour Cool

Mark:

TABLES & CHAIRS

Maximize 10C + 20T subject to:
2C + T <= 120
C + T <= 72
C + 4T <216>= 0
T >= 0

The solution space is a pentagon bounded by the lines described above.
The revenue function will be maximized at a vertex of the solution space.
The vertices (intersections of the boundary lines) and their associated revenues are (C,T):
(0,0): $0
(60,0): $600
(48,24): $960
(24,48): $1200
(0,54): $1080

Revenue is maximized at $1200 with 24 chairs and 48 tables. Cool

To check; it will take 24*10 + 48*5 = 480 minutes of saw time. There are 600 minutes available so the saws will be idle 20% of the time.

It will take 24*5 + 48*5 = 360 minutes of lathe time which is exactly what we have.

It will take 24*5 + 48*20 = 1080 minutes of sanding machine time which is exactly what we have.

Thanks guys you are the best! Very Happy

Slippy; who I believe was crowned Miss Chicago 1937 writes, "LADDERS Riding high?"

If you are looking for a leg up; come over to see me sometime!(Don't forget your jodhpurs!) Razz

cwaentaitirdons Question

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Tryagain wrote:

cwaentaitirdons

weird incantations Laughing

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is there a proof that a maximum will occur on the vertex of the figure and not inside or on an edge?

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gee thoh, you sound so mathematical Laughing

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Thoh, is this what you seek:

Often, the simplest way to solve "ax2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or you just don't feel like factoring.

So, while factoring may not always be successful, the Quadratic Formula can always find the solution.

The Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers. The Formula is derived from the process of completing the square, and is formally stated as:
For ax^2 + bx + c = 0, the value of x is given by:

x = - b (minus or plus) Sqrt b^2 - 4ac
.2a

Note that, for the Formula to work, you must have "(quadratic) = 0". Note also that the "2a" at the bottom of the Formula is underneath everything above, not just the square root.

And don't forget that it's a "2a" under there, not just a "2"! And make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back" , and you'll mess yourself up.

And remember that "b2" means "the square of ALL of b, including the sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive.

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THOH
(The Handsome Other One)

do join us on the ship hunni. Smile

oh....sorry....thought this was the socialising thread

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thoh13:

Check out the simplex algorithm. From wikipedia:

Quote:

A linear programming problem consists of a collection of linear inequalities on a number of real variables and a given linear function (on these real variables) which is to be maximized or minimized. Further details are given in the linear programming article.

In geometric terms we are considering a closed convex polytope, P, defined by intersecting a number of half-spaces in n-dimensional Euclidean space; each half-space is the area which lies on one side of a hyperplane. If the objective is to maximise a linear functional L(x), consider the hyperplanes H(c) defined by L(x) = c; as c increases, these form a parallel family. If the problem is well-posed, we want to find the largest value of c such that H(c) intersects P (if there is no such largest value of c, this isn't a reasonable question for optimization as it stands). In this case we can show that the optimum value of c is attained on the boundary of P. Methods for finding this optimum point on P work in several ways: some attempt to improve a possible point by moving through the interior of P (so-called interior point methods); others start and remain on the boundary searching for an optimum.

The simplex algorithm follows this latter methodology. The idea is to move along the facets of P in search of the optimum, from point to point. Note that, unless the optimum occurs on an edge or face that is parallel to H, the optimum will be unique and occur at a vertex of the polytope. If an optimum is found on an edge or face that is parallel to H then the optimum is not unique and can be obtained at any point on the edge or face. Since the simplex algorithm is concerned only with finding a single optimal point (even if other equally-optimal points exist), it is possible to look solely at moves skirting the edge of a simplex, ignoring the interior. The algorithm specifies how this is to be done.

It's the linearity of the constraints and revenue function that guarantee a maximum at a vertex.

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thoh13 wrote:

is there a proof that a maximum will occur on the vertex of the figure and not inside or on an edge?

Or a minimum---one of the vertices of the pentagon is no chairs and no tables---and in the realm of manufactured profits you can't get much lower than none, that is if you ignore capital and operating costs.

Markr linear ;linear programming is much less tedious than my differential optimization solution---even though they both skinned the same cat.

Rap

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TTH:

cwaentaitirdons = weird incantations Cool

How the heck did you manage that! Razz

Mark and Rap arrive in a restaurant independently. Each arrives a random time between 5pm and 6pm, distributed uniformly (no moment in this range is any more likely for arrival than another).

What is the probability they arrived within 10 minutes of each other Question

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