Optimization problem

[dr|ft]

In a beehive, each cell is a regular prism, open at one end with a trihedral angle at the other end, it is believed that bees form their cells in such a way as to minimize the surface area for a given volume, thus using the least amount of wax in cell construction examination of these cells has shown that the measure of the apex angle is amazingly consistent based on the geometry of the cell, it can be shown that the surface area is given by A = 6sh -1.5s^2 cot X + ((3 square root 3s^2)/2sinX) where s is the length of sides of the hexagon and h is the height , and they are constants.

Use the optimisation theory to determine the minimum surface area of the cell in terms of s and h, verify the answer by sketching the Surface A function on graphmatica for spec values of s and h. ( these values may be arbitarily chosen) use the first derivative test to theoretically prove that X produces a min value.

 

Thx

[DQW]

Well ? How do you intend to go about this homework problem ? Any ideas ?

[dr|ft]

=] deriving then letting A = 0 ? somethign along those lines

[Dave]

Just thought I'd point out that the derivative of a constant is zero.

[dr|ft]

So we want to derive A = 6sh -1.5s^2 cot X + ((3 square root 3s^2)/2sinX)....

 

6sh would go to 0.... -1.5s^2 cotX would go to... derivative of cotX is -csc^2 X so it goes to 1.5x^2csc^2 X

 

((3 square root 3s^2)/2sinX) would go to - 1.5rt3s^2 csc X . cot X ???

 

making it..

 

A' = 1.5s^2 csc^2 X - 1.5root3s^2 csc X . cot X

 

????

 

EDIT

 

checking that last bit of the equation on maple

> diff((3*sqrt(3*s^2))/(2*sin(x)),x);

 

 

3 .root(3).root(s^2).cos(x)

- -------------------------

 

2 sin(x)^2

 

I'm assuming this is correct, if so could someone please show me the steps in getting that

 

EDIT:

ok for the last bit ((3 square root 3s^2)/2sinX) i think you use the quotient rule so lets call 3sqrt(3s^2) 'f' and 2sinX 'g' ..... the rule is g.[df/dx] - f.[dg/dx] all over g^2

so that will equate to [2sinX.0 - 3sqrt(3s^2).2cosX] / 4sin^2X

or (-3sqrt(3s^2).2cosX)/(4sin^2X)

[DQW]

A' = 1.5s^2 csc^2 X - 1.5root3s^2 csc X . cot X

That's correct !

 

EDIT

 

checking that last bit of the equation on maple

> diff((3*sqrt(3*s^2))/(2*sin(x)),x);

 

 

3 .root(3).root(s^2).cos(x)

- -------------------------

 

2 sin(x)^2

 

I'm assuming this is correct, if so could someone please show me the steps in getting that

 

EDIT:

ok for the last bit ((3 square root 3s^2)/2sinX) i think you use the quotient rule so lets call 3sqrt(3s^2) 'f' and 2sinX 'g' ..... the rule is g.[df/dx] - f.[dg/dx] all over g^2

so that will equate to [2sinX.0 - 3sqrt(3s^2).2cosX] / 4sin^2X

or (-3sqrt(3s^2).2cosX)/(4sin^2X)

Notice that all of these are giving you the same result, since

 

[math] \frac {cos(x)}{sin^2(x)} = \frac {cos(x)}{sin(x)} \frac {1}{sin(x)} = cot(x) cosec(x) [/math]

 

Go ahead; you're doing fine. Equate this to 0 and find X.

[DQW]

One little detail. I'm pretty sure the last term in the original equation is [imath]3 \sqrt{3} s^2 /2sinX[/imath] , not [imath]3 \sqrt{3 s^2} /2sinX [/imath]. The second option is not even dimensionally correct.

[dr|ft]

yeah that's right DQW, it's [3.root(3).s^2]/[2sinX]....so now i have

0=1.5s^2.csc^2X - 1.5root(3).s^2.cscX.cotX and i want to find X....

I would start by splitting the equation to look like

 

1.5s^2.csc^2X = 1.5root(3).s^2.cscX.cotX

 

divide both sides by 1.5 and s^2

 

csc^2X = root(3)cscX.cotX

 

seperate the X's by dividing RHS by cscX.cotX

 

csc^2X / cscX.cotX = root(3)

 

cancel the csc's leaving

 

cosecX / cotX = root(3)

 

hmmmmmmm hit a wall

 

EDIT:

I can plug tanX/sinX = root(3) in my graphics calc and solve for X to get 54.7356 which seems OK..... btw i got tanX/sinX by going cosecX = 1/sinX and cotX = 1/tanX. Anyone have any ideas for manually equating?

[DQW]

Two things :

 

1. You must state why you are justified in canceling csc(x) - because you can never have csc(x) = 0 for any x.

 

2. csc/cot = tan/sin = (sin/cos)*(1/sin) = 1/cos = sec(x)

[dr|ft]

can't i cancel csc because its csc^2X / cscX.cotX = root(3)

which is the same as

 

cosecX/cosecX * cosecX/cotX = root(3) ?

 

EDIT:

If i solve secX=root(3) X comes out to be 0.99 or something which couldn't possibly be correct

 

DOUBLE EDIT:

Wait no i put it in as 1/cosX=root(3) and it came out as 54.7356 again

[dr|ft]

Is my next step to go back to A' = 1.5s^2 csc^2 X - 1.5root(3)s^2 csc X . cot X

 

let A=0, X=54.7356 and solve for s... so it would be

 

0 = 1.5s^2csc^254.7356 - 1.5root(3)s^2.csc54.7356

 

hmmm that doesn't look right it seems the s' will cancel out

[DQW]

No, you don't solve for s. s can take any value...which leads you to the next part of the question. Plug in X = 54.74 in the equation for the surface area, A, and thus get an expression for the minimal surface are in terms of s and h. Then you plot A(s,h) for different values of s and h.

 

PS : You can cancel off X/X in all cases EXCEPT when X=0, since "0/0" is not defined. Fortunately, csc(x) is never equal to zero.

[dr|ft]

Wait my quesiton only wanted me to find Surface Area with respect to s and h so i just go back to the original equation and sub in 54.7356 for theta, is that what you guys got?

[DQW]

Yes, that's correct.

[dr|ft]

you got 54.7356 degrees? cool. My question also asks me to 'theoretically' prove that theta produces a minimum value using the 'first derivative test', is that what we've already done or is there more to it?