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**.

Once we have such a **simplified equation,** we can then **optimize i**t 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.

## 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:

\[Â 10 \ foot \ \times \ 10Â \ foot \ \times \ 5 \ footÂ \]

## Numerical Result

\[Â 10 \ foot \ \times \ 10Â \ foot \ \times \ 5 \ footÂ \]

## 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:

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