Question

 An open box is to be made out of a 8-inch by 16-inch piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. Find the dimensions of the resulting box that has the largest volume.

Answer 1

Let $x=$ the side os a square, $0\leq x\leq4.$

The volume of the box is

$$V=V(x)=x(8-2x)(16-2x)$$

$$=4(x^3-12x^2+32x), 0\leq x\leq 4$$

Find the first derivative with respect to $x$

$$V'(x)=4(3x^2-24x+32)$$

Find the critical number(s)

$$V'(x)=0=>4(3x^2-24x+32)=0$$

$$3x^2-24x+32=0$$

$$D=(-24)^2-4(3)(32)=192$$

$$x=\dfrac{24\pm \sqrt{192}}{2(3)}=4\pm\dfrac{4\sqrt{3}}{3}$$

Since $0\leq x\leq 4,$ we take $x=4-\dfrac{4\sqrt{3}}{3}$

If $0<x<4-\dfrac{4\sqrt{3}}{3}, V'(x)>0, V(x)$ increases.

If $4-\dfrac{4\sqrt{3}}{3}<x<4, V'(x)<0, V(x)$ decreases.

The function $V(x)$ has a local maximum at $x=4-\dfrac{4\sqrt{3}}{3}.$

Since the function $V(x)$ has the only extremum on $[0, 4],$ then the function $V(x)$ has the absolute maximum on $[0, 4]$ at $x=4-\dfrac{4\sqrt{3}}{3}.$

The dimensions of the resulting box that has the largest volume are

$$\dfrac{8\sqrt{3}}{3} \ in\times (8+\dfrac{8\sqrt{3}}{3})\ in\times (4-\dfrac{4\sqrt{3}}{3})\ in$$