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worty [1.4K]
3 years ago
11

HELP ME PLEASEEE !!!!!!!

Mathematics
1 answer:
Sindrei [870]3 years ago
5 0
And it is also too small
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You must design a closed rectangular box of width w, length l and height h, whose volume is 504 cm3. The sides of the box cost 3
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Dimensions will be

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A closed box has length = l cm

width of the box = w cm

height of the box = h cm

Volume of the rectangular box = lwh

504 = lwh

h=\frac{504}{lw}

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Total cost of the box C = 3024\frac{(l+w)}{lw} + 8lw

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\frac{dC}{dl}=3024(-l^{-2}+0)+8w=0

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\frac{378}{l^{2}}=w ---------(1)

\frac{dC}{dw}=3024(-w^{-2})+8l=0

\frac{3024}{w^{2}}=8l

\frac{378}{w^{2}}=l

w^{2}=\frac{378}{l}-------(2)

Now place the value of w from equation (1) to equation (2)

(\frac{378}{l^{2}})^{2}=\frac{378}{l}

\frac{(378)^{2} }{l^{4}}=\frac{378}{l}

l³ = 378

l = ∛378 = 7.23 cm

From equation (2)

w^{2}=\frac{378}{7.23}

w^{2}=52.28

w = 7.23 cm

As lwh = 504 cm³

(7.23)²h = 504

h=\frac{504}{(7.23)^{2}}

h = 9.64 cm                        

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