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

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

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.)
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Answer 2

#h=2.82876, r = 1.06078#

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#
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Answer 3

To find the dimensions that minimize the cost of material, you typically need a specific context or problem statement. The general approach involves setting up a cost function based on the dimensions of the object or structure in question, then using calculus techniques such as differentiation to find the critical points where the cost function reaches a minimum. These critical points can then be analyzed to determine the optimal dimensions. If you provide the specific context or problem statement, I can assist you further in solving it.

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Answer 4

To find the dimensions that will minimize the cost of the material, you need to provide the specific context or details of the problem, such as the shape of the object being constructed, the cost function for the material, and any constraints or limitations. Once those details are provided, mathematical techniques such as calculus optimization can be applied to determine the dimensions that minimize the cost.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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