Question

A closed rectangular box whose length is double its width has a total surface area of 600 cm². Find the dimensions of the box with maximum volume.

Answer 1

The surface area of a rectangular prism with the width $w$, length $l$, and height $h$ h is given by:

$S=2(wl+wh+lh)$

Substitute $S=600, ~\text{and}~~l=2w$ into the above equation:

$600=2(w \cdot 2w +wh+2wh)$

Simplify the equation:

$600=2(2w^{2} +wh+2wh)$

Find $h$ in terms of $w$:

Divide both sides of the equation by $2$:

$300=2w^{2} +wh+2wh$

Combine like terms:

$300=2w^{2} +3wh$

Move the expression $3wh$ to the left-hand side and change its sign:

$300-3wh=2w^{2}$

Move the expression to the left-hand side and change its sign:

$-3wh=2w^2-300$

Divide both sides of the equation by $-3w$:

$h=\frac{2w^2-300}{-3w}$

The volume of the rectangular box is given by:

$V=w \cdot l \cdot h$

Substitute $l=2w$ and $h=\frac{2w^2-300}{-3w}$ into the equation:

$V=w \cdot 2w \cdot \frac{2w^2-300}{-3w}$

Simplify the equation:

$V=-\frac{4}{3}w^{3}+200w$

To find the maximum volume, differentiate $V$ with respect to $w$

$\frac{dv}{dt}=-4w^{2}+200$

Set $\frac{dv}{dt}=0$

$-4w^{2}+200=0$

Move the expression $200$ to the left-hand side and change its sign:

$-4w^{2}$ $=-200$

Divide both sides by $-4$ :

$w^2=50$

Take the square root for both sides:

$w=\pm 5\sqrt{2}$

Since there is no negative lengths, exclude the negative sign:

$w=5\sqrt{2}$

$l=2w=2 \times 5\sqrt{2}=10\sqrt{2}$

$h=\frac{2(5\sqrt{2})^2-300}{-3(5\sqrt{2})}$

Using a calculator:

$h=\frac{20\sqrt{2}}{3}$