Lets make a picture of our problem:
where h denotes the height of the box.
We know that the volume of a rectangular prism is
[tex]\begin{gathered} V=(4x)(x)(h) \\ V=4x^2h \end{gathered}[/tex]
Since the volume must be 8 cubic centimeters, we have
[tex]4x^2h=48[/tex]
Then, the height function is equal to
[tex]h=\frac{48}{4x^2}=\frac{12}{x^2}[/tex]
On the other hand, the function cost C is given by
[tex]C=1.80A_{\text{bottom}}+1.80A_{\text{top}}+2\times3.60A_{\text{side}1}+2\times3.60A_{\text{side}2}[/tex]
that is,
[tex]\begin{gathered} C=1.80\times4x^2+1.80\times4x^2+3.60(8xh+2xh) \\ C=3.60\times4x^2+3.60\times10xh \end{gathered}[/tex]
which gives
[tex]C=3.60(4x^2+10xh)[/tex]
By substituting the height result from above, we have
[tex]C=3.60(4x^2+10x(\frac{12}{x^2}))[/tex]
which gives
[tex]C=3.60(4x^2+\frac{120}{x})[/tex]
Now, in order to find minum cost, we need to find the first derivative of the function cost and equate it to zero. It yields,
[tex]\frac{dC}{dx}=3.60(8x-\frac{120}{x^2})=0[/tex]
which is equivalent to
[tex]\begin{gathered} 8x-\frac{120}{x^2}=0 \\ \text{then} \\ 8x=\frac{120}{x^2} \end{gathered}[/tex]
by moving x squared to the left hand side and the number 8 to the right hand side, we have
[tex]\begin{gathered} x^3=\frac{120}{8} \\ x^3=15 \\ \text{then} \\ x=\sqrt[3]{15} \\ x=2.4662 \end{gathered}[/tex]
Therefore, by substituting this value in the height function, we get
[tex]h=\frac{12}{2.4662^2}=1.9729[/tex]
therefore, by rounding to the neastest hundredth, the height which minimize the cost is equal to 1.97 cm
