Serious Calc help please...I'm lost and I have a huge test tomorrow that I am studying for?

Question

Problem: A tin can is made with thicker tin on top and bottom than on the side. The tin used for the top and bottom costs 20% more than the material for the side. FInd the dimensions of the can that minimize cost if the can must have a volume of 5 cu. in

Basically I have a whole section of optimization problems like this...mostly these are going to be on my test. Not only can you show me how to solve it, but can you also please help explain what different steps you take in order to solve either finding maxima or minima. Thank you.

Answer 1

V = πr²h
A (top & bottom) = πr²
End Area = 2πr²
Lateral Area = 2πrh
Cost of tube = 2πrhC, C being cost per unit area
Cost of ends = 2πr²C*1.2
Total cost, P = 2πC(1.2r² + rh)
h = V/πr², so
P = 2πC(1.2r² + rV/πr²)
P = 2πC(1.2r² + V/πr)
dP/dr = 2πC(2.4r - V/πr²) = 0 for max or min, so
2.4r = V/πr²
r³ = V/2.4π
r³ = 5/2.4π
r = 0.872"
h = 5/π0.872²
h = 2.093"

Check:
C(2.4*π0.872² + 10/0.872) = 17.201050C
C(2.4*π0.87² + 10/0.87) = 17.201444C (+0.000394)
C(2.4*π0.874² + 10/0.874) = 17.201137C (+0.000087)

Answer 2

OK. These max/min problems always break down into a set of problems
1) Get a formula for what you are trying to minimize or maximize. 2) Take a derivative of the function that you determine.
3) Solve the deriviative for values that will make it zero
4) Check that the second derivative isn't also zero at that point
5) Work out what that minimum value is
6) Check your boundary conditions

So, let's have a look a the problem that you posed.
We need a cost function. Let's let c be the cost of the side material. It will probably end up cancelling out a the end.
Let's let r be the radius of the can and h be the height

Do you see that we have a cost for the tin of:
2*pi*r^2*1.2*c+pi*2*r*h*c?

Collect a few like terms and you see that the cost for a can is
2*pi*c*(1.2*r^2+r*h)

OK so what is the volume of the can? That's
pi*r^2*h

To minimize the cost per volume we need a formula for that:
2*pi*c*(1.2*r^2+r*h)/(pi*r^2*h)

We can simplify that a little bit to
2*pi*c*(1.2/h+1/r)

SO, pick a variable to eliminate by plugging in the known volume, differentiate, solve for zero, check for plausiblity, and go on to the next one.

Answer 3

V = volume of a cylinder = Pi*h*r^2 = 5 in^3
As = surface area of side = 2*Pi*r*h
Ae = surface area of both ends = 2*Pi*r^2
We assume cost is proportional to area
C = total cost = [1.2*Ae + As] * k
where k is cost per unit area of the thin tin for side
but we don't need to know this, so we can "normalize"
the equation with a new variable c = C/k
c = 1.2*Ae + As = 2*Pi*[1.2*r^2 + h*r]
now we need to minimize c subject to the constraint of V
we can use V solved for h and substitute into c
h = 5/(Pi*r^2)
c = 2*Pi*[1.2*r^2 + 5/(Pi*r)]
solve dc/dr =0
dc/dr = 4.8*Pi*r - 10/r^2 = 0
r = [10/(4.8*Pi)]^(1/3) = .872 in
h = 5/(Pi*r^2) = 2.09 in