Mai wants to make an open top box by cutting out corners of a square piece of cardboard and folding up the sides. The cardboard

Question

3 months ago

9

is 10 cm by 10 cm. The volume V(x) in cubic cm of the open top box is a function of the side length x in cm of the square cutouts

1 answer:

Inessa [12.5K] · Answer 1 · 2026-02-18T16:47:33+03:00

Answer:

$V(x)=(4x^{3}-40x^{2}+100x)\ cm^3$

The variable x lies within the interval of all positive real numbers less than 5 cm.

Detailed solution:

Problem statement:

Determine the volume of the open-topped box as a function of the side length x (in centimeters) of the square cutouts.

Refer to the provided diagram for clarity.

Define:

x → length in centimeters of each square cutout side

The volume of the box with open top can be written as:

$V=LWH$

Given this, we have:

$L=(10-2x)\ cm$

$W=(10-2x)\ cm$

$H=x)\ cm$

By substitution:

$V(x)=(10-2x)(10-2x)x\\\\V(x)=(100-40x+4x^{2})x\\\\V(x)=(4x^{3}-40x^{2}+100x)\ cm^3$

Determine the domain of x:

Because:

$(10-2x) > 0\\10> 2x\\ 5 > x\\x < 5\ cm$

Therefore:

Domain is the interval (0,5)

That means all real numbers strictly greater than zero and less than 5 cm are valid for x.

Hence, the volume V as a function of x is:

$V(x)=(4x^{3}-40x^{2}+100x)\ cm^3$