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A box of volume 72 m3 with a square bottom and no top is constructed out of two different materials. the cost of the bottom is $40/m2 and the cost of the sides is $30/m2 . find the dimensions of the box that minimize total cost.

Respuesta :

apologiabiology
square bottom, so L=W
yah

V=HL²

surface area (minus the top) is
SA=2H(2L)+LW
SA=4HL+L²
great
given V=72
72=HL²
solve for H
72/(L²)=H

umm
sorry, I'm all over the place
cost=30(4)HL+40L² or
cost=120HL+40L²
subsitute for H

cost=120(72/(L²))L+40L²
cost=8640/L+40L²
find the minimum using your calculator or take the derivitive
min at L=3∛4≈4.7622031559046

anyway, sub back

72/L²=H
72/((3∛4)²)=H
2∛4=H

the dimentions of the box are
height is 2∛4 or about 3.1748m
length and width are 3∛4 or about 4.7622m

The dimensions of the box that minimize total cost are height[tex](h)=3.17m[/tex] , width[tex](w)[/tex] and length[tex](l)[/tex]  are equal to [tex]4.76m[/tex] .

Given Volume of the box is [tex]72[/tex] [tex]m^{3}[/tex] with no top.

And the cost of the bottom is [tex]40 USD/m^{2}[/tex].

For the sides, it is [tex]30USD/m^{2}[/tex]

The bottom of the box is square, so [tex]l=w[/tex] and let [tex]h[/tex] is the height of the box.

So Volume

[tex]V=hl^{2}\\72=hl^{2} \\h=\dfrac{72}{l^{2} }[/tex]

Now Surface Area[tex](S)[/tex] of the box without a top is

[tex]S=4hl+l^{2}[/tex]

So,

[tex]cost=4\times30hl +40l^{2} \\120hl+40l^{2}[/tex]

Putting [tex]h[/tex],

[tex]cost=120\times(72/l^{2} )l+40l^{2}\\cost=\dfrac{8640}{l^{2} } +40l^{2}[/tex]

To find minimum cost, derivate the [tex]cost[/tex] by using the calculator at

[tex]l=3\sqrt[3]{4}\\ l=4.7622031559046 m[/tex]

[tex]h=\dfrac{72}{(3\sqrt[3]{4} )^{2} }\\h=3.17m[/tex]

Hence the dimensions of the box are

[tex]l=w\\=4.76m.[/tex]

and height  [tex]h=3.17m[/tex]

Know more about the volume of the box here:

https://brainly.com/question/11168779?referrer=searchResults

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