Maximizing Volume of Open-Top Box

We are given a rectangular box with an open top. Let the dimensions be length ($l$), width ($w$), and height ($h$).

• The length is twice the width: $l = 2w$.
• The outer surface area ($A$) is 12 m$^2$. The surface area consists of the base and four sides (since it's open top).
  $A = (\text{base area}) + 2 (\text{front/back area}) + 2 (\text{side area})$
  $A = (l \times w) + 2 (l \times h) + 2 (w \times h)$
• Substitute $l=2w$:
  $A = (2w \times w) + 2 (2w \times h) + 2 (w \times h)$
  $A = 2w^2 + 4wh + 2wh = 2w^2 + 6wh$
• Using the given area $A=12$:
  $12 = 2w^2 + 6wh$
• The volume ($V$) of the box is $V = l \times w \times h$.
  Substitute $l=2w$: $V = (2w) \times w \times h = 2w^2h$.

Expressing Volume in Terms of Width

We need to express the volume $V$ using only one variable (e.g., $w$). From the surface area equation, we can express $h$ in terms of $w$:

• $12 = 2w^2 + 6wh$
  $6wh = 12 - 2w^2$
  $h = \frac{12 - 2w^2}{6w} = \frac{6 - w^2}{3w}$
• Substitute this expression for $h$ into the volume formula:
  $V(w) = 2w^2 \left( \frac{6 - w^2}{3w} \right)$
  $V(w) = \frac{2w(6 - w^2)}{3} = \frac{12w - 2w^3}{3}$

Finding Dimensions for Maximum Volume

To maximize the volume $V(w)$, we find the derivative with respect to $w$ and set it to zero.

• $\frac{dV}{dw} = \frac{d}{dw} \left( \frac{12w - 2w^3}{3} \right) = \frac{1}{3} (12 - 6w^2)$
• Set $\frac{dV}{dw} = 0$:
  $\frac{1}{3} (12 - 6w^2) = 0$
  $12 - 6w^2 = 0$
  $6w^2 = 12$
  $w^2 = 2$
  $w = \sqrt{2}$ (since width must be positive)
• We can confirm this is a maximum using the second derivative test:
  $\frac{d^2V}{dw^2} = \frac{1}{3} (-12w) = -4w$. For $w=\sqrt{2}$, $\frac{d^2V}{dw^2} = -4\sqrt{2} < 0$, indicating a maximum.

Calculating the Ratio of Length to Height

Now we find the dimensions $l$ and $h$ when $w = \sqrt{2}$ and calculate their ratio.

• $w = \sqrt{2}$ m
• $l = 2w = 2\sqrt{2}$ m
• $h = \frac{6 - w^2}{3w} = \frac{6 - (\sqrt{2})^2}{3\sqrt{2}} = \frac{6 - 2}{3\sqrt{2}} = \frac{4}{3\sqrt{2}}$ m
• Ratio $\frac{l}{h}$:
  $\frac{l}{h} = \frac{2\sqrt{2}}{\frac{4}{3\sqrt{2}}} = 2\sqrt{2} \times \frac{3\sqrt{2}}{4} = \frac{6 \times (\sqrt{2})^2}{4} = \frac{6 \times 2}{4} = \frac{12}{4} = 3$

The ratio of length to height is 3.