eyeofthetiger14
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if 10 800 cm2 of material is available to make a box with a square base and an open top find the largest possible volume of the box.

Sagot :

calculista

Let

x--------> the length side of the square base of the box

y-------> the height of the box

we know that

The surface area of the box is equal to

[tex]SA=4xy+x^{2}[/tex]

[tex]SA=10,800\ cm^{2}[/tex]

so

[tex]10,800=4xy+x^{2}[/tex]

[tex]y=(10,800-x^{2})/(4x)[/tex]

[tex]y=(2,700-0.25x^{2})/(x)[/tex] --------> equation A

the volume of the box is equal to

[tex]V=x^{2}y[/tex] --------> equation B

Substitute the equation A in the equation B

[tex]V=x^{2}*(2,700-0.25x^{2})/(x)[/tex]

[tex]V=x*(2,700-0.25x^{2})[/tex]

[tex]V=(2,700x-0.25x^{3})[/tex]

using a graphing tool

see the attached figure

For [tex]x=60\ cm[/tex]

[tex]Volume=108,000\ cm^{3}[/tex]

the point [tex](60,108,000)[/tex] is a maximum of the function

Find the dimensions of the box

[tex]x=60\ cm[/tex]

Find the value of y

[tex]V=x^{2}y[/tex]

[tex]y=V/x^{2}[/tex]

[tex]y=108,000/60^{2}[/tex]

[tex]y=30\ cm[/tex]

The dimensions of the box are

[tex]60\ cm*60\ cm*30\ cm[/tex]

The largest possible volume of the box is

[tex]108,000\ cm^{3}[/tex]

View image calculista

The largest possible volume of the box is [tex]\boxed{108000{\text{ c}}{{\text{m}}^3}}.[/tex]

Further explanation:

Given:

The area of the material is [tex]10800{\text{ c}}{{\text{m}}^2}.[/tex]

Explanation:

Consider the base length of the square box as “x”.

Consider the height of the box as “y”.

The surface area of the open box can be expressed as follows,

[tex]\boxed{{\text{Surface Area}} = 4xy + {x^2}}[/tex]

The surface area of the box is [tex]10800{\text{ c}}{{\text{m}}^2}.[/tex]

[tex]\begin{aligned}4xy + {x^2}&= 10800\\4xy&= 10800 - {x^2}\\y&= \frac{{10800 - {x^2}}}{{4x}}\\y&=\frac{{2700 - 0.25{x^2}}}{x}\\\end{aligned}[/tex]

The volume of the box can be expressed as follows,

[tex]\begin{aligned}V&= {x^2}y\\&= {x^2}\times \left({\frac{{2700 - 0.25{x^2}}}{x}} \right)\\&= \left( x \right)\times \left({2700 - 0.25{x^2}} \right)\\&= 2700x - 0.25{x^3}\\\end{aligned}[/tex]

Differentiate the volume with respect to “x”.

[tex]\begin{aligned}\frac{{dV}}{{dx}}&= \frac{d}{{dx}}\left({2700x - 0.25{x^3}}\right)\\&= 2700 - 0.75{x^2}\\\end{aligned}[/tex]

Substitute 0 for [tex]\dfrac{{dV}}{{dx}}[/tex] in above equation to obtain the value of x.

[tex]\begin{aligned}0&= 2700 - 0.75{x^2}\\0.75{x^2} &= 2700\\{x^2}&= \frac{{2700}}{{0.75}}\\{x^2}&= 3600\\x&= 60\\\end{aligned}[/tex]

The side of the base is [tex]60{\text{ cm}}.[/tex]

The height of the box can be obtained as follows,

[tex]\begin{aligned}y&= \frac{{2700 - 0.25{{\left( {60} \right)}^2}}}{{60}}\\&=\frac{{2700 - 900}}{{60}}\\&=\frac{{1800}}{{60}}\\&=30\\\end{aligned}[/tex]

The height of the box is [tex]y = 30{\text{ cm}}.[/tex]

The volume of the box can be calculated as follows,

[tex]\begin{aligned}V&={\left(60}\right)^2}\times\left({30}\right)\\&=3600\times30\\&=108000 \\\end{aligned}[/tex]

The largest possible volume of the boxis  [tex]\boxed{108000{\text{ c}}{{\text{m}}^3}}[/tex].

Learn more:

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Application of Derivatives

Keywords: square, box, material, square base, volume of the box, largest, open from the top derivative, surface area.

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