Question

Find the dimensions that will minimize the cost of the material?

A cylindrical container that has a capacity of `10\text( m)^3` is to be produced.

• The top and bottom of the container are to be made of a material that costs `$20` per square meter,
• while the side of that container is to be made of a material costing `$15` per square meter.

Find the dimensions that will minimize the cost of the material.

Answer 1

`h=2.82876, r = 1.06078`

Explanation:

Calling `c_1=20` and `c_2=15` the total cost is

`C=2(pi r^2)c_1+(2pirh)c_2`.

Here `r` is the base radius and `h` is the side height.

The volume is given by

`V=pir^2h=V_0=10`.

Now, the problem can be stated as

`min_(h,r)C(h,r)` restricted to `V(h,r)=V_0`

Using lagrange multipliers it reads

`L(h,r,lambda)=C(h,r)+lambda (V(h,r)-V_0)`

with stationary points given by

`grad L = vec 0` or

#{(2 c_2 pi r + pi r^2 lambda= 0), (2 c_2 h pi+ 4 c_1 pi r + 2 h pi r lambda = 0), (h pi r^2 - V_0 =0):}#

solving for `h,r,lambda` we get

`h = root(3)((2c_1/c_2)^2V_0/pi), r= root(3)(c_2/c_1V_0/(2pi)),lambda = -2root(3)((2pic_1c_2^2)/(V_0))` or

`h=2.82876, r = 1.06078` with associated cost `C=424.214`

We know that is a minimum because

`(C@V)(r)=(2 (c_1 pi r^3 + c_2 V_0))/r` and
`(d^2)/(dr^2)(C@V)(r)=(4 (c_1 pi r^3 + c_2 V_0))/r^3`

and for the found solution has the value

`(d^2)/(dr^2)(C@V)(r)=240pi > 0`

Answer 2

For students who have not yet learned calculus of two variables, here is the single variable solution.

Explanation:

A (right circular) cylinder has two variables,

height, `h`, and
radius, `r`.

The top and bottom cost `pir^2(20)` $. SInce there are two of these, they add `40pir^2` $ to the cost.

The cost of the side of the cylinder will be `2pirh(15) = 30pirh` $

The total cost is `40pir^2 + 30pirh`.

In order to make this a function of a single variable, we need an equation with both `h` and `r` in it.

The volume of a (right circular) cylinder is `V = pi r^2h` and we want the volume to be `10 " " m^3`.

`pi r^2h = 10`, so

`h = 10/(pir^2)`.

The cost can now be written as a function of `r` alone.

`C(r) = 40pir^2 + 30pirh = 40pir^2 + 30pir(10/(pir^2))`

`= 40pir^2 + 30pir(10/(pir^2))`

`= 40pir^2 + 300/r`

Domain is `r > 0`

We want to minimize `C`, so we'll find the derivative, then critical number(s) and then test the critical number(s).

`C'(r) = 80pir-300/r^2 = (80pir^3-300)/r^2`

The only real valued critical number is `r = root(3)(15/(4pi))`.

Using the first or second derivative test verifies that `C` is minimum at this critical number.

Second derivative test:
`C''(x) = 80pi+600/r^3`

So `C''(root(3)(15/(4pi))) = 80pi+600/(15/(4pi)) > 0`

Finally, the question asks for the dimensions, so we have

`r = root(3)(15/(4pi))` and

`h = 10/(pir^2) = 10/(pi(root(3)(15/(4pi)))^2)`

`= 10/pi (root(3)((4pi)/15))^2` (Rewrite further as desired.)