Fencing optimization problem for calculus
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Fencing optimization problem for calculus

[From: ] [author: ] [Date: 12-03-20] [Hit: ]
or about 21.21 mOur width is 30√2, or about 42.43 mThe total amount of the fencing material that you need should then be 4l + 2w.lets solve for that f using both of our equations containing f to make sure:f = 4l + 2wf = 4(15√2) + 2(30√2)f = 60√2 + 60√2f = 120√2 msecond equation:f = 4l + 1800/lf = 4(15√2) + 1800/(15√2)f = 60√2 + 120/√2f = 60√2 + 120√2 / 2f = 60√2 + 60√2f = 120√2 mWe get the same f both times.So our l and w must be correct.......

But what we do want to know is what w is. so go back to equation 1 that had length and width equal to the area, and solve for w:

900 = lw
900 = (15√2)w
w = 900 / (15√2)
w = 60 / √2

simplify:

w = 60√2 / 2
w = 30√2

So our length is 15√2, or about 21.21 m
Our width is 30√2, or about 42.43 m

The total amount of the fencing material that you need should then be 4l + 2w. let's solve for that f using both of our equations containing f to make sure:

f = 4l + 2w
f = 4(15√2) + 2(30√2)
f = 60√2 + 60√2
f = 120√2 m

second equation:

f = 4l + 1800/l
f = 4(15√2) + 1800/(15√2)
f = 60√2 + 120/√2
f = 60√2 + 120√2 / 2
f = 60√2 + 60√2
f = 120√2 m

We get the same f both times. So our l and w must be correct.

-
x = width
y = length

x * y = 900
2y + 4x = P

x = 900/y

2y + 4x =>
2y + 3600/y

2y + 3600/y = P
2 - 3600/y^2 = dP/dy

dP/dy = 0

0 = 2 - 3600/y^2
3600/y^2 = 2
3600 = 2y^2
y^2 = 1800
y = sqrt(100 * 18)
y = sqrt(100) * sqrt(9) * sqrt(2)
y = 10 * 3 * sqrt(2)
y = 30 * sqrt(2)

x * y = 900
x * 30 * sqrt(2) = 900
x * sqrt(2) = 30
x = 30 / sqrt(2)
x = 30 * sqrt(2) / 2
x = 15 * sqrt(2)
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keywords: calculus,Fencing,problem,optimization,for,Fencing optimization problem for calculus
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