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3 years ago

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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

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:

Bumek [7]3 years ago

5 0

Answer:

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

The domain for x is all real numbers greater than zero and less than 5 com

Step-by-step explanation:

<em><u>The question is</u></em>

What is the volume of the open top box as a function of the side length x in cm of the square cutouts?

see the attached figure to better understand the problem

Let

x -----> the side length in cm of the square cutouts

we know that

The volume of the open top box is

$V=LWH$

we have

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

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

$H=x)\ cm$

substitute

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

Find the domain for x

we know that

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

so

The domain is the interval (0,5)

The domain is all real numbers greater than zero and less than 5 cm

therefore

The volume of the open top box as a function of the side length x in cm of the square cutouts is

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

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