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Wewaii [24]
3 years ago
13

You have $7000 with which to build a rectangular enclosure with fencing. The fencing material costs $30 per meter. You also want

to have two partitions across the width of the enclosure, so that there will be three separated spaces in the enclosure. The material for the partitions costs $25 per meter. What is the maximum area you can achieve for the enclosure
Mathematics
1 answer:
Arlecino [84]3 years ago
4 0

Answer:

1857.12 square meters

Step-by-step explanation:

Let L be the length of the rectangle and 'W' be the width

Perimeter = 2L + 2W + 2W

The fencing material costs $30 per meter.

The material for the partitions costs $25 per meter

cost=30(2L + 2W) +25( 2W)\\60L+60W+50W\\60L +110W

7000=60L+110W

Solve for L

7000=60L+110W\\7000-110W= 60L\\L=\frac{700}{6} -\frac{11}{6} W

Area = length times width

( \frac{700}{6} -\frac{11}{6} W)(W)\\A(W)=\frac{700}{6}W -\frac{11}{6} W^2

Now take derivative and set it =0

A(W)=\frac{700}{6}W -\frac{11}{6} W^2\\A'(W)=\frac{700}{6} -\frac{22}{6} W

set the derivative =0  and solve for W

0=\frac{700}{6} -\frac{22}{6} W\\\frac{700}{6}=\frac{22}{6} W\\W= 31.8

So width = 31.8 that gives maximum area

L=\frac{700}{6} -\frac{11}{6} (31.8)=58.4

Area = length  \cdot width = 31.8 \cdot 58.4= 1857.12 square meter

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Answer:

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Step-by-step explanation:

Given

See attachment

Required

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Quantity = 2 * 3ft = 6ft

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Rate = \frac{Price}{Quantity}

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Rate = \frac{\$3.40}{12\ ft}

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ehidna [41]

Answer:

The answer is "\frac{25}{3}".

Step-by-step explanation:

please find the complete question in the attached file.

a_0=\frac{1}{10}\int^{5}_{x=-5} x^2 dx\\

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
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yaroslaw [1]

Answer: 24.625

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3 years ago
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