Help with maximum/minimum problems??

Question

The cost per square metre of the sides of an open-topped cylindrical tank is twice the cost per square metre of the bottom. Find the most economical proportions for a tank of given volume.
**I dont know how to incorporate the costs...?
**p.s. the answer should be 1:2

Answer 1

The total cost is side cost + bottom cost and it must be minimum. Let´s call total cost = w and side cost = c

w = sidearea x 2c + bottomarea x c
w = 2(pi)rh x 2c + (pi)r^2 x c
w = 4(pi)crh + (pi)cr^2 it is a function of two variables r and h
The partial derivative must be zero in the minimum

dw/dr = 4(pi)ch + 2(pi)cr
dw/dh = 4(pi)cr + 0

Now we have a linear system: 4(pi)ch + 2(pi)cr =0
4(pi)cr + 0 = 0

From the second equation we get r = c/(4(pi))
Putting this in the first equation we get h= c/(8(pi))

These are the values of r and h for minimum cost.

The proportion is r/c = 1/2

Answer 2

Are you familiar with the usual routines in this sort of question?write the formulas for surface area and volume:

S = 2*pi*r*h + pi*r^2
V = pi*(r^2)*h. Ask what is constant? It's volume, so write

h = V/(pi*r^2) remembering V and pi are constants.

The only twist, as you suggest, is that it's about minimum cost, not minimum area, so instead of that first equation, let the cost per square unit of the bottom be 1 money unit (we could call it c dollars, but that just clutters up the equations with another unnecessary constant.) Then the sides cost 2 money units per square unit, and the cost is y money units where
y = 2*2*pi*r*h + pi *r^2. I think you can do the rest, but if not, scroll down to see the rest of my solution(After I edit and re-post). I can also be reached at [e-mail]

No need, I see vahucel has already finished it. But I'll complete it below anyway without referring to "partial derivatives"


































































y = 4*pi*r*V/(pi*r^2) + pi*r^2
dy/dr = -4V/r^2 + 2*pi*r which is zero when

r^3 =2V/pi = 2(r^2)h
whence r = 2h, i.e. r:h = 1:2

Of course, we should also show that it is in fact minimum.
y" = 8V/r^3 + 2pi, and since r and V are positive, y" is positive. Therefore the stationary value is a local minimum.