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Evgesh-ka [11]
3 years ago
7

Engineers are designing a box-shaped aquarium with a square bottom and an open top. The aquarium must hold 256 ft^3 of water. Wh

at dimensions should they use to create an acceptable aquarium with the least amount of glass?
Mathematics
1 answer:
Andrej [43]3 years ago
7 0
This is a calculus problem (in optimization).  Let the aquarium dimensions be L, W and H.  Then L*W*H = 256 ft^3.  But W=L here, so   W^2*H = 256 ft^3.

Write an expression for the area of the aquarium's glass sides and bottom.

                                                       256 ft^3                             256 ft^3
A= W^2 + 4(W*H) = W^2 + 4*W*(------------- ) = W^2 + 4*-------------------
                                                          W^2                                      W

So now we have A(W), the area as a function of W alone.

We want to minimize this area.  To do this, differentiate A(W) with respect to W and set the result = to 0.  We want to determine the "critical numbers."

dA/dW = 2W - 1024*W^(-2) = 0
                        1024
Then 2W = ---------------
                       W^2

2W^3 = 1024, so W^3 = 512, and  W = third root of 512 = 8

If W = 8 ft, then L = 8 ft also.     Since  L*W*H = 256 ft^3,

L*W*H = 256 ft^3   = (8 ft)^2*H = 256 ft^3.  Then H = 4

The acquarium dimensions are 8 by 8 by 4 feet.  This keeps the area of the aquarium to a minimum.
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Step-by-step explanation:

Using the points :

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m = (80 - 62) / (6 - 0)

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Family size at $0.20 per ounce is our correct answer.

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Family size:

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We see that the family-size box of cereals is the lowest per ounce.

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The form of the partial fraction decomposition is \frac{At + B}{(t^2 + 4)} + \frac{Ct + D}{(t^2 + 2)} + \frac{Et + F}{(t^2 + 2)^2}

<h3>How to decompose the function?</h3>

The fraction is given as:

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To decompose the above, we make use of the following rule:

\frac{px^n + qx^m +....+ r}{(x^2 + a)(x^2 + b)^2} = \frac{Ax + B}{(x^2 + a)} + \frac{Cx + D}{(x^2 + b)} + \frac{Ex + F}{(x^2 + b)^2}

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