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Your iron works has contracted to design and build a 500 cubic foot, square-based, open-top, rectangular steel holding tank for a paper company. The tank is made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible. What dimensions do you tell the shop to use?

The aim of this question is to optimize the surface area of the box.

To solve this question, we first find some constraints and try to generate an equation of surface area that only has one variable.

Solid

Solid

Once we have such a simplified equation, we can then optimize it by the differentiation method. We first find the first derivative of the surface area equation. Then we equate it to zero to find the local minima. Once we have this minimum value, we apply the constraints to find the final dimensions of the box.

First derivative

First derivative

2Nd derivative

2Nd derivative

Expert Answer

The total surface area of the box can be calculated using the following formula:

\[ \text{ Surface Area of the Box } \ = \ S \ = \ 4 \times ( \text{ Rectangular Sides } ) \ + \ \text{ Square Base } \]

Let us assume that:

\[ \text{ Length and width of square base } \ = \ x \]

Also since:

\[ \text{ Rectangular Sides } \ = \ x \times h \]

\[ \text{ Square Base } \ = \ x \times x \ = \ x^{ 2 }\]

Substituting these values in the above equation:

\[ S \ = \ 4 \times ( x \times h ) \ + \ x^{ 2 } \ … \ … \ … \ ( 1 ) \]

The volume of such a box can be calculated using the following formula:

\[ V \ = \ x \times x \times h \]

\[ \Rightarrow V \ = \ x^{ 2 } \times h \]

Given that:

\[ V \ =\ 500 \ square \ foot \]

The above equation becomes:

\[ 500 \ cubic \ foot \ = \ x^{ 2 } \times h \]

\[ \Rightarrow h \ = \ \dfrac{ 500 }{ x^{ 2 } } \ … \ … \ … \ ( 2 ) \]

Substituting the value of h from equation (1) in equation (2):

\[ S \ = \ 4 \times ( x \times \dfrac{ 500 }{ x^{ 2 } } ) \ + \ x^{ 2 } \]

\[ \Rightarrow S \ = \ \dfrac{ 2000 }{ x } \ + \ x^{ 2 } \]

Taking derivative:

\[ S’ \ = \ – \dfrac{ 2000 }{ x^{ 2 } } \ + \ 2x \]

Minimizing S:

\[ 0 \ = \ – \dfrac{ 2000 }{ x^{ 2 } } \ + \ 2x \]

\[ \Rightarrow \dfrac{ 2000 }{ x^{ 2 } } \ = \ 2x \]

\[ \Rightarrow 2000 \ = \ 2x^{ 3 } \]

\[ \Rightarrow 1000 \ = \ x^{ 3 } \]

\[ \Rightarrow ( 10 )^{ 3 } \ = \ x^{ 3 } \]

\[ \Rightarrow x \ = \ 10 \ foot \]

Substituting this value in equation (2):

\[ h \ = \ \dfrac{ 500 }{ ( 10 )^{ 2 } } \]

\[ \Rightarrow h \ = \ \dfrac{ 500 }{ 100 } \]

\[ \Rightarrow h \ = \ 5 \ foot \]

Hence, the minimum dimensions that will use the minimum surface area or minimum mass of metal will be as follows:

\[  10 \ foot \ \times \ 10 \ foot \ \times \ 5 \ foot  \]

Numerical Result

\[  10 \ foot \ \times \ 10 \ foot \ \times \ 5 \ foot  \]

Example

If the mass per square foot of the metal sheets used is 5 kg, then what will be the weight of the final product after manufacturing?

Recall equation (1):

\[ S \ = \ 4 \times ( x \times h ) \ + \ x^{ 2 } \]

Substituting values:

\[ S \ = \ 4 \times ( 10 \times 5 ) \ + \ ( 5 )^{ 2 } \ = \ 200 \ + \ 25 \ = \ 225 \ square \ foot \]

The weight of the metal can be calculated with the following formula:

\[ m \ = \ S \times \text{ mass per square foot } \ = \ 225 \times 5 \ = \ 1125 \ kg \]

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