mcminty wrote:
mmm, so..
Defining the volume as V = (pi)r2h = 16(pi) and Surface Area as A = 2(pi)r2h + 2(pi)rh
We need to minimise function f = 2(pi)r2h + 2(pi)rh;
Subject to function g = (pi)r2h - 16(pi) = 0
In otherwords, we'll find the radius (and height) that gives the minimum surface area while also giving 16(pi) cm^3 volume.
The Lagrangian function: L = f - µg (I used µ in place of usually using lambda, if I go by my notes)
Thus L = 2(pi)r2h + 2(pi)rh - µ*[(pi)r2h - 16(pi)]
Now, you need to partially differentiate function L, and set to zero. In other words, come up with ∂L/∂r = 0 and ∂L/∂h = 0.
From here, you'll have to find r and h in terms of µ. You should be able to read one of the values off straight away, and sub it into the other equation to get the other value. (ie. the ∂L/∂h = (whatever) = 0 should only have r terms in it, so you'll be able to simply find r = whatever*µ... and then sub that into the second equation)
Once you have r and h in terms of µ, sub them back into the initial conditions for volume, ie. (pi)r2h = 16(pi) and it will yield an actual value for µ. Then finally use this µ value and substitute this back into the equations of r and h in terms of µ, to get the required radius and height values.
whoa whoa whoa, i've done problems in the past like this and we never had to use Lagrangian functions (i don't even know what they are) or mu. all that was done was substitution, then differentiation.
i don't really want to use a Lagrangian function because this is summer work and it's supposed to be a test of what we know...and i do know it, i just can't remember the final piece.
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