Question

I am doing an equation trying to figure out a formula for the volume of a box and I am so lost I will include a picture

Answer 1

The volume of any rectangular box is expressed as:

`\text{Volume}=\text{length}*\text{breadth}* height`

Now, for the box that will be formed from the figure shown in the question, we will have:

length = 37 - 2x

breadth = 37 - 2x

height = x

Thus, we have that:

`\begin{gathered} \text{Volume}=\text{length}*\text{breadth}* height \\ \Rightarrow\text{Volume}=(37-2x)*(37-2x)* x \end{gathered}`

We now simplify the above as:

`\begin{gathered} \text{Volume}=(37-2x)*(37-2x)* x \\ \Rightarrow\text{Volume}=(1369-148x+4x^2)* x \\ \Rightarrow\text{Volume}=1369x-148x^2+4x^3 \\ \Rightarrow\text{ V(x)}=1369x-148x^2+4x^3 \end{gathered}`

Now that we have obtained the expression for the volume of the box, we now have to find the value of x that maximizes it.

This is done as follows:

Method

- Differentiate the function V(x) with respect to x, and equate to zero as follows:

`\begin{gathered} \Rightarrow V^1\text{(x)}=1369-296x^{}+12x^2 \\ \text{Equating to zero:} \\ 1369-296x^{}+12x^2=0 \\ \text{The roots of the equation are:} \\ \Rightarrow x=6.167\text{ and x = }18.5 \end{gathered}`

Now we have to find the second derivative of V(x) in order to confirm which value of x makes the function V(x) a maximum

Thus:

`\begin{gathered} \Rightarrow V^(11)\text{(x)}=-296^{}+24x^{} \\ \text{when x = 6.167} \\ \Rightarrow V^(11)\text{(6.167)}=-296^{}+24(6.167)=-296+148.008=-148 \\ \text{when x = }18.5 \\ \Rightarrow V^(11)\text{(18.5)}=-296^{}+24(18.5)=-296+444=148 \end{gathered}`

Now since the second derivative is a negative number when x = 6.167, we now know for sure that it is that value of x that maximizes the function V(x), and not x = 18.5.

Thus, we can conclude that the value of x that maximizes the volume of the box is:

x = 6.17 inches (to 2 decimal places)

If we had been asked to find the value of x that minimizes the volume, the answer will have been x = 18.5, because this value of x made the second derivative of V(x) positive.

Now, the maximum volume of the box is obtained by simply substituting the value of x that maximizes the function into the original expression for V(x), as follows:

`\begin{gathered} V(x)=1369x-148x^2+4x^3 \\ \text{when x= 6.167} \\ \Rightarrow\text{ V(6.167)}=1369(6.167)-148(6.167)^2+4(6.167)^3 \\ \Rightarrow\text{ V(6.167)}=8442.623-5628.720+938.171 \\ \Rightarrow\text{ V(6.167)}=3752.074in^3 \\ \Rightarrow\text{ V(6.167)}=3752.07in^3\text{ (to 2 decimal places)} \end{gathered}`