How do you find the cost of materials for the cheapest such container given a rectangular storage container with an open top is to have a volume of #10m^3# and the length of its base is twice the width, and the base costs $10 per square meter and material for the sides costs $6 per square meter?

Answer 1
Let #x# and #y# be the lenght and width of the base, and #z# be the height.
Knowing that the lenght is twice the width means that #x=2y#
Knowing that the volume is fixed at #10m^3#, means that #x*y*z=10#
From the previous relation, we can express #x# in terms of #y#, obtaining #x*y*z=10 \rightarrow 2y*y*z=10 \rightarrow 2y^2*z=10#, and from this equality we can obtain #z# as a function of #y#: #z=10/{2y^2}= 5/y^2#
Expressing all the three variables in terms of one is important, because now we have a problem which depends on a single variable: the area of the basis is #x*y=2y^2#, and the lateral surface surface is given by #2(x+y)z=6y*5/y^2=30/y#
Since the base costs #$10# per square meters, and the lateral surface costs #$6# per square meters, the total cost is given by #10*2y^2 + 6*30/y=20y^2 + 180/y#.
So, #c(y)=20y^2 + 180/y# is the function that we want to minimize, to minimize the cost. Let's study its first derivative, and find the values for which it equals zero:
#c'(y)=40y - 180/y^2=0 \iff 40y = 180/y^2 \iff y^3=9/2# So, if you choose #y# as the cube root of #9/2#, you'll minimize the function, and thus the cost.
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Answer 2

To find the cost of materials for the cheapest such container, we need to minimize the total cost, considering the cost of the base and the cost of the sides.

Given:

  • Volume of the container = 10 m³
  • Length of the base = twice the width
  • Cost of base material = $10 per square meter
  • Cost of side material = $6 per square meter

Let's denote:

  • Length of the base as ( L )
  • Width of the base as ( W )
  • Height of the container as ( H )

From the given information, we know that the volume of the container (( V )) is given by: [ V = L \times W \times H ]

Given that ( L = 2W ), we can rewrite the volume equation as: [ V = 2W \times W \times H = 2W^2H ]

Since the volume (( V )) is given as 10 m³, we have: [ 10 = 2W^2H ]

We need to minimize the cost function ( C ), which is the sum of the cost of the base and the cost of the sides: [ C = 10WL + 6(2LH + 2WH) ]

Substituting ( L = 2W ) into the cost function, we get: [ C = 10(2W)W + 6\left(2(2W)H + 2W(H)\right) ] [ C = 20W^2 + 6(4WH + 2WH) ] [ C = 20W^2 + 6(6WH) ] [ C = 20W^2 + 36WH ]

Now, we can express ( H ) in terms of ( W ) using the volume equation: [ H = \frac{10}{2W^2} = \frac{5}{W^2} ]

Substituting this into the cost function, we get: [ C = 20W^2 + 36W\left(\frac{5}{W^2}\right) ] [ C = 20W^2 + \frac{180}{W} ]

To minimize ( C ), we take its derivative with respect to ( W ) and set it equal to zero: [ \frac{dC}{dW} = 40W - \frac{180}{W^2} = 0 ]

Solving this equation gives us the value of ( W ), which we can then use to find ( L ) and ( H ), and ultimately calculate the minimum cost.

Once we find the values of ( W ), ( L ), and ( H ), we substitute them into the cost function ( C ) to find the minimum cost of materials for the container.

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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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