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.

2 Answers
Nov 21, 2016

#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#

Nov 21, 2016

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.)