Question

A box is made from material that has a dimension 24 by 48. The box is to be form by cutting squares from each of the four edges of the material. What are the dimensions of the box such that the volume is maximum

Answer 1

(Refer to image for diagram.)

Volume is length*width*height
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 $24=2*12$, $x$
⇒$x=12-4\sqrt{3}$ 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 $8\sqrt{3}$×$(24+8\sqrt{3})$×$(12-4\sqrt{3})$