Question

1. find the dimensions of the cardboard box (shaped like a rectangular solid) with no top which maximizes volume among all boxes made from 48 feet² of cardboard.

Explanation:

Step1: Let the length be $l$, width be $w$ and height be $h$.

The surface - area formula for a box without a top is $S = lw+2lh + 2wh=48$, so $lw+2h(l + w)=48$, and $h=\frac{48 - lw}{2(l + w)}$. The volume formula is $V=lwh$. Substitute $h$ into the volume formula: $V=\frac{lwh(48 - lw)}{2(l + w)}$. For simplicity, assume the base is a square, i.e., $l = w$. Then the surface - area formula becomes $S=x^{2}+4xh = 48$, so $h=\frac{48 - x^{2}}{4x}$.

Step2: Substitute $h$ into the volume formula.

The volume $V=x\cdot x\cdot h=x^{2}h$. Substituting $h=\frac{48 - x^{2}}{4x}$ into $V$, we get $V=\frac{x(48 - x^{2})}{4}=\frac{48x - x^{3}}{4}$.

Step3: Take the derivative of $V$ with respect to $x$.

Using the power rule, if $V=\frac{48x - x^{3}}{4}=12x-\frac{1}{4}x^{3}$, then $V^\prime=12-\frac{3}{4}x^{2}$.

Step4: Set the derivative equal to zero to find critical points.

$12-\frac{3}{4}x^{2}=0$. Rearranging gives $\frac{3}{4}x^{2}=12$, then $x^{2}=16$, so $x = 4$ (we ignore $x=-4$ since length cannot be negative).

Step5: Find the value of $h$.

Substitute $x = 4$ into $h=\frac{48 - x^{2}}{4x}$. So $h=\frac{48-16}{4\times4}=\frac{32}{16}=2$.

Answer:

The dimensions of the box are length $l = 4$ feet, width $w = 4$ feet and height $h = 2$ feet.