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Anestetic [448]
3 years ago
5

PLEASE HELP!!!!!!!!!!!!!!

Mathematics
1 answer:
MaRussiya [10]3 years ago
6 0
The answer is the top answer

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2 years ago
An open box is to be constructed so that the length of the base is 3 times larger than the width of the base. If the cost to con
natta225 [31]

Answer:

Step-by-step explanation:

Let assume that the volume = 88 cubic feet

Then:

L = 3w --- (1)volume = l \times w \times h \\ \\ 88 =  (3w)   \times w  \times h \\ \\ 3w^2 h= 88 \\ \\ h = \dfrac{88}{3w^2}--- (2) \\ \\

The construction cost now is:

C = 4(l \times w) + 2 ( w  \times h) + 2(l \times h) \\ \\ C = 4(3w^2) + 2(w  \times \dfrac{88}{3w^2}) + 2(3w \times \dfrac{88}{3w^2}) \\\\ C = 12w^2 + \dfrac{176}{3w} + \dfrac{176}{w}

Now, to determine the minimum cost:

\dfrac{dC}{dw}= 0  \\ \\ \implies 24 w - \dfrac{176}{3w^2}- \dfrac{176}{w^2}=0 \\ \\ 24 w ^3 = \dfrac{176}{w^2}(\dfrac{1}{3}+1) \\ \\ 24 w ^3 = \dfrac{176(4)}{3} \\ \\ w^3 = \dfrac{88}{3(3)}

w = \dfrac{88}{3(3)}^{1/3} \ feet

Now;

length = 3 ( \dfrac{88}{3(3)})^{1/3} \ feet

height = ( 88})^{1/3} (3)^{1/3} \ feet

6 0
3 years ago
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