If 2400 square centimeters of material is available to make a box with a square base and an open top, how do you find the largest possible volume of the box?

Question

If 2400 square centimeters of material is available to make a box with a square base and an open top, how do you find the largest possible volume of the box?

1 Answer

Impact of this question

34383 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-03T00:37:49

The dimension of the box I found are:
$28.3 \times 28.3 \times 14.1 c m$ $28.3xx28.3xx14.1cm$ for a volume of $11, 292.5 c m^{3}$ $11,292.5cm^3$ :

Explanation:

The surface $S$ $S$ of a box with open top is the sum of the surfaces of a square (base, of side $a$ $a$ ) and 4 rectangles (base $a$ $a$ and height $h$ $h$ ):
$S = a^{2} + 4 a \cdot h$ $S=a^2+4a*h$ (1)
while the volume $V$ $V$ will be the area of the base times the height, or:
$V = a^{2} \cdot h$ $V=a^2*h$ (2)

We get $h = \frac{S - a^{2}}{4 a}$ $h=(S-a^2)/(4a)$ from (1) put it into (2):

$V = a^{2} \frac{S - a^{2}}{4 a} = \frac{1}{4 a} (S a^{2} - a^{4}) = \frac{S}{4} a - \frac{a^{3}}{4}$ $V=a^2(S-a^2)/(4a)=1/(4a)(Sa^2-a^4)=S/4a-a^3/4$
maximize this volume deriving with respect to $a$ $a$ and setting it equal to zero:
$\frac{d V}{d a} = \frac{S}{4} - \frac{3}{4} a^{2} = 0$ $(dV)/(da)=S/4-3/4a^2=0$
$\frac{3}{4} a^{2} = \frac{S}{4}$ $3/4a^2=S/4$
$a^{2} = \frac{S}{3}$ $a^2=S/3$
$a = \pm \sqrt{\frac{2400}{3}} = \pm 28.3 c m$ $a=+-sqrt(2400/3)=+-28.3cm$

We use $a = + 28.3 c m$ $a=+28.3cm$ that in (1) gives us:
$h = \frac{2400 - {28.3}^{2}}{4 \cdot 28.3} = 14.1 c m$ $h=(2400-28.3^2)/(4*28.3)=14.1cm$

Giving a volume of:
$V = a^{2} \cdot h = 11, 292.5 c m^{3}$ $V=a^2*h=11,292.5cm^3$

Science

Math

Humanities

... and beyond

If 2400 square centimeters of material is available to make a box with a square base and an open top, how do you find the largest possible volume of the box?

1 Answer

Explanation:

Related questions

Impact of this question