(Refer to image for diagram.)
Volume is lengthwidthheight
So \(V(x)=x(48-2x)(24-2x)\)
⇒\(V(x)=4x(24-x)(12-x)\)
⇒\(V(x)=4x(288-36x+x^{2})\)
⇒\(V(x)=4(288x-36x^{2}+x^{3})\)
Differentiate: \(V'(x)=4(288-72x+3x^{2})\)
For maximum, set \(V'(x)=0\)
⇒\(4(288-72x+3x^{2})=0\)
⇒\(96-24x+x^{2}=0\)
⇒\(x=\frac{24+-\sqrt{(-24)^2-4(96)}}{2}\)
⇒\(x=\frac{24+-\sqrt{576-384}}{2}\)
⇒\(x=\frac{24+-\sqrt{192}}{2}\)
⇒\(x=\frac{24+-8\sqrt{3}}{2}\)
⇒\(x=12+-4\sqrt{3}\)
Since theshorter side of the original rectangle is \(24=2*12\), \(x<12\)
⇒\(x=12-4\sqrt{3}\) is the height
For the longer side of the base:
\(48-2x=48-2(12-4\sqrt{3})=48-24+8\sqrt{3}=24+8\sqrt{3}\)
For the shorter side of the base:
\(24-2x=24-2(12-4\sqrt{3})=24-24+8\sqrt{3}=8\sqrt{3}\)
The dimensions are \(8\sqrt{3}\)×\((24+8\sqrt{3})\)×\((12-4\sqrt{3})\)
