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