Optimization Question?
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A farmer has 6000 m of fencing and wishes to create a rectangular filed subdivided into four congruent plots of land. Determine the dimensions of the each plot if the area to be enclosed is a maximum.
A: 375 m by 600m
My Attempt:
Fencing:
6000 = 3x + 3y
(6000-3x)/3 = y
Area
A = xy
A = x(6000-3x)/3
A' = (18,000 - 18x)/9
0 = 18,000 - 18x
x = 1000
why do i keep getting this wrong? Thank you in advance!
calculus
asked Jan 26 '14 at 8:02
JessicaJessica
42631326
$endgroup$
add a comment |
$begingroup$
A farmer has 6000 m of fencing and wishes to create a rectangular filed subdivided into four congruent plots of land. Determine the dimensions of the each plot if the area to be enclosed is a maximum.
A: 375 m by 600m
My Attempt:
Fencing:
6000 = 3x + 3y
(6000-3x)/3 = y
Area
A = xy
A = x(6000-3x)/3
A' = (18,000 - 18x)/9
0 = 18,000 - 18x
x = 1000
why do i keep getting this wrong? Thank you in advance!
calculus
asked Jan 26 '14 at 8:02
JessicaJessica
42631326
$endgroup$
$begingroup$
What are 375 m and 600 m ? They don't appear anywhere.
$endgroup$
– Claude Leibovici
Jan 26 '14 at 8:14
$begingroup$
If $ x $ and $ y $ represent the dimensions of each plot, how many lengths of fence and how many widths are needed for the four parallel rectangles? (The answers are not 3 and 3...) Is there a picture that goes with this -- it matters whether the four smaller rectangles are parallel or arranged 2-by-2.
$endgroup$
– colormegone
Jan 26 '14 at 8:18
add a comment |
$begingroup$
A farmer has 6000 m of fencing and wishes to create a rectangular filed subdivided into four congruent plots of land. Determine the dimensions of the each plot if the area to be enclosed is a maximum.
A: 375 m by 600m
My Attempt:
Fencing:
6000 = 3x + 3y
(6000-3x)/3 = y
Area
A = xy
A = x(6000-3x)/3
A' = (18,000 - 18x)/9
0 = 18,000 - 18x
x = 1000
why do i keep getting this wrong? Thank you in advance!
calculus
asked Jan 26 '14 at 8:02
JessicaJessica
42631326
$endgroup$
A farmer has 6000 m of fencing and wishes to create a rectangular filed subdivided into four congruent plots of land. Determine the dimensions of the each plot if the area to be enclosed is a maximum.
A: 375 m by 600m
My Attempt:
Fencing:
6000 = 3x + 3y
(6000-3x)/3 = y
Area
A = xy
A = x(6000-3x)/3
A' = (18,000 - 18x)/9
0 = 18,000 - 18x
x = 1000
why do i keep getting this wrong? Thank you in advance!
calculus
calculus
asked Jan 26 '14 at 8:02
JessicaJessica
42631326
asked Jan 26 '14 at 8:02
JessicaJessica
42631326
asked Jan 26 '14 at 8:02
JessicaJessica
42631326
asked Jan 26 '14 at 8:02
JessicaJessica
42631326
asked Jan 26 '14 at 8:02
JessicaJessica
42631326
42631326
$begingroup$
What are 375 m and 600 m ? They don't appear anywhere.
$endgroup$
– Claude Leibovici
Jan 26 '14 at 8:14
$begingroup$
If $ x $ and $ y $ represent the dimensions of each plot, how many lengths of fence and how many widths are needed for the four parallel rectangles? (The answers are not 3 and 3...) Is there a picture that goes with this -- it matters whether the four smaller rectangles are parallel or arranged 2-by-2.
$endgroup$
– colormegone
Jan 26 '14 at 8:18
add a comment |
$begingroup$
What are 375 m and 600 m ? They don't appear anywhere.
$endgroup$
– Claude Leibovici
Jan 26 '14 at 8:14
$begingroup$
If $ x $ and $ y $ represent the dimensions of each plot, how many lengths of fence and how many widths are needed for the four parallel rectangles? (The answers are not 3 and 3...) Is there a picture that goes with this -- it matters whether the four smaller rectangles are parallel or arranged 2-by-2.
$endgroup$
– colormegone
Jan 26 '14 at 8:18
$begingroup$
What are 375 m and 600 m ? They don't appear anywhere.
$endgroup$
– Claude Leibovici
Jan 26 '14 at 8:14
$begingroup$
What are 375 m and 600 m ? They don't appear anywhere.
$endgroup$
– Claude Leibovici
Jan 26 '14 at 8:14
$begingroup$
If $ x $ and $ y $ represent the dimensions of each plot, how many lengths of fence and how many widths are needed for the four parallel rectangles? (The answers are not 3 and 3...) Is there a picture that goes with this -- it matters whether the four smaller rectangles are parallel or arranged 2-by-2.
$endgroup$
– colormegone
Jan 26 '14 at 8:18
$begingroup$
If $ x $ and $ y $ represent the dimensions of each plot, how many lengths of fence and how many widths are needed for the four parallel rectangles? (The answers are not 3 and 3...) Is there a picture that goes with this -- it matters whether the four smaller rectangles are parallel or arranged 2-by-2.
$endgroup$
– colormegone
Jan 26 '14 at 8:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A rectangular region may be divided into four congruent pieces in many ways. Regardless of how they're divided, each of their areas is maximized if the total enclosed area is maximized. (A and 4A are simultaneously maximized.)
$2 times 2$: Let $x,y$ be the lengths of the enclosed region. The enclosed area is $A = xy$. The fencing for the perimeter and the internal fences is $P=3x+3y=6000 text{ m}$. We see that $y = 2000 text{ m} -x$ so $A = x(2000 text{ m} - x)$ which is a parabola with maximum at $x=1000 text{ m}$ so $y=1000 text{ m}$. Each plot is then $500 text{ m} times 500 text{ m}$
$1 times 4$: Let $x,y$ be the lengths of the enclosed region, with $x$ being the length of the sides common to neighboring subplots. Then again $A = xy$ and now $P = 5x+2y = 6000 text{ m}$. This time, $y = 3000 text{ m} - (5/2)x$, so $A = x(3000 text{ m} - (5/2)x)$. This parabola takes a maximum when $x=600 text{ m}$ so $y=1500 text{ m}$. Each plot is then $600 text{ m} times 375 text{ m}$
Your error is that you're subdividing incorrectly.
Edit: Divided $y$ among the lengths of each of the four subparcels.
Edit: Corrected a stupid units transposition.
answered Jan 26 '14 at 8:43
Eric TowersEric Towers
32.8k22370
$endgroup$
$begingroup$
how did you get 5x + 2y? Where does the 5 and the 2 come from? Also, how did you get 375 from 1500?
$endgroup$
– Jessica
Jan 26 '14 at 14:22
1
$begingroup$
@Jessica: Draw the picture of the $1 times 4$ division of the plot. Then count segments with the same lengths. Did you draw a picture? That's step one in attacking word problems...
$endgroup$
– Eric Towers
Jan 27 '14 at 6:26
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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active
oldest
votes
$begingroup$
A rectangular region may be divided into four congruent pieces in many ways. Regardless of how they're divided, each of their areas is maximized if the total enclosed area is maximized. (A and 4A are simultaneously maximized.)
$2 times 2$: Let $x,y$ be the lengths of the enclosed region. The enclosed area is $A = xy$. The fencing for the perimeter and the internal fences is $P=3x+3y=6000 text{ m}$. We see that $y = 2000 text{ m} -x$ so $A = x(2000 text{ m} - x)$ which is a parabola with maximum at $x=1000 text{ m}$ so $y=1000 text{ m}$. Each plot is then $500 text{ m} times 500 text{ m}$
$1 times 4$: Let $x,y$ be the lengths of the enclosed region, with $x$ being the length of the sides common to neighboring subplots. Then again $A = xy$ and now $P = 5x+2y = 6000 text{ m}$. This time, $y = 3000 text{ m} - (5/2)x$, so $A = x(3000 text{ m} - (5/2)x)$. This parabola takes a maximum when $x=600 text{ m}$ so $y=1500 text{ m}$. Each plot is then $600 text{ m} times 375 text{ m}$
Your error is that you're subdividing incorrectly.
Edit: Divided $y$ among the lengths of each of the four subparcels.
Edit: Corrected a stupid units transposition.
answered Jan 26 '14 at 8:43
Eric TowersEric Towers
32.8k22370
$endgroup$
$begingroup$
how did you get 5x + 2y? Where does the 5 and the 2 come from? Also, how did you get 375 from 1500?
$endgroup$
– Jessica
Jan 26 '14 at 14:22
1
$begingroup$
@Jessica: Draw the picture of the $1 times 4$ division of the plot. Then count segments with the same lengths. Did you draw a picture? That's step one in attacking word problems...
$endgroup$
– Eric Towers
Jan 27 '14 at 6:26
add a comment |
$begingroup$
A rectangular region may be divided into four congruent pieces in many ways. Regardless of how they're divided, each of their areas is maximized if the total enclosed area is maximized. (A and 4A are simultaneously maximized.)
$2 times 2$: Let $x,y$ be the lengths of the enclosed region. The enclosed area is $A = xy$. The fencing for the perimeter and the internal fences is $P=3x+3y=6000 text{ m}$. We see that $y = 2000 text{ m} -x$ so $A = x(2000 text{ m} - x)$ which is a parabola with maximum at $x=1000 text{ m}$ so $y=1000 text{ m}$. Each plot is then $500 text{ m} times 500 text{ m}$
$1 times 4$: Let $x,y$ be the lengths of the enclosed region, with $x$ being the length of the sides common to neighboring subplots. Then again $A = xy$ and now $P = 5x+2y = 6000 text{ m}$. This time, $y = 3000 text{ m} - (5/2)x$, so $A = x(3000 text{ m} - (5/2)x)$. This parabola takes a maximum when $x=600 text{ m}$ so $y=1500 text{ m}$. Each plot is then $600 text{ m} times 375 text{ m}$
Your error is that you're subdividing incorrectly.
Edit: Divided $y$ among the lengths of each of the four subparcels.
Edit: Corrected a stupid units transposition.
answered Jan 26 '14 at 8:43
Eric TowersEric Towers
32.8k22370
$endgroup$
$begingroup$
how did you get 5x + 2y? Where does the 5 and the 2 come from? Also, how did you get 375 from 1500?
$endgroup$
– Jessica
Jan 26 '14 at 14:22
1
$begingroup$
@Jessica: Draw the picture of the $1 times 4$ division of the plot. Then count segments with the same lengths. Did you draw a picture? That's step one in attacking word problems...
$endgroup$
– Eric Towers
Jan 27 '14 at 6:26
add a comment |
$begingroup$
A rectangular region may be divided into four congruent pieces in many ways. Regardless of how they're divided, each of their areas is maximized if the total enclosed area is maximized. (A and 4A are simultaneously maximized.)
$2 times 2$: Let $x,y$ be the lengths of the enclosed region. The enclosed area is $A = xy$. The fencing for the perimeter and the internal fences is $P=3x+3y=6000 text{ m}$. We see that $y = 2000 text{ m} -x$ so $A = x(2000 text{ m} - x)$ which is a parabola with maximum at $x=1000 text{ m}$ so $y=1000 text{ m}$. Each plot is then $500 text{ m} times 500 text{ m}$
$1 times 4$: Let $x,y$ be the lengths of the enclosed region, with $x$ being the length of the sides common to neighboring subplots. Then again $A = xy$ and now $P = 5x+2y = 6000 text{ m}$. This time, $y = 3000 text{ m} - (5/2)x$, so $A = x(3000 text{ m} - (5/2)x)$. This parabola takes a maximum when $x=600 text{ m}$ so $y=1500 text{ m}$. Each plot is then $600 text{ m} times 375 text{ m}$
Your error is that you're subdividing incorrectly.
Edit: Divided $y$ among the lengths of each of the four subparcels.
Edit: Corrected a stupid units transposition.
answered Jan 26 '14 at 8:43
Eric TowersEric Towers
32.8k22370
$endgroup$
A rectangular region may be divided into four congruent pieces in many ways. Regardless of how they're divided, each of their areas is maximized if the total enclosed area is maximized. (A and 4A are simultaneously maximized.)
$2 times 2$: Let $x,y$ be the lengths of the enclosed region. The enclosed area is $A = xy$. The fencing for the perimeter and the internal fences is $P=3x+3y=6000 text{ m}$. We see that $y = 2000 text{ m} -x$ so $A = x(2000 text{ m} - x)$ which is a parabola with maximum at $x=1000 text{ m}$ so $y=1000 text{ m}$. Each plot is then $500 text{ m} times 500 text{ m}$
$1 times 4$: Let $x,y$ be the lengths of the enclosed region, with $x$ being the length of the sides common to neighboring subplots. Then again $A = xy$ and now $P = 5x+2y = 6000 text{ m}$. This time, $y = 3000 text{ m} - (5/2)x$, so $A = x(3000 text{ m} - (5/2)x)$. This parabola takes a maximum when $x=600 text{ m}$ so $y=1500 text{ m}$. Each plot is then $600 text{ m} times 375 text{ m}$
Your error is that you're subdividing incorrectly.
Edit: Divided $y$ among the lengths of each of the four subparcels.
Edit: Corrected a stupid units transposition.
answered Jan 26 '14 at 8:43
Eric TowersEric Towers
32.8k22370
answered Jan 26 '14 at 8:43
Eric TowersEric Towers
32.8k22370
answered Jan 26 '14 at 8:43
Eric TowersEric Towers
32.8k22370
answered Jan 26 '14 at 8:43
Eric TowersEric Towers
32.8k22370
32.8k22370
$begingroup$
how did you get 5x + 2y? Where does the 5 and the 2 come from? Also, how did you get 375 from 1500?
$endgroup$
– Jessica
Jan 26 '14 at 14:22
1
$begingroup$
@Jessica: Draw the picture of the $1 times 4$ division of the plot. Then count segments with the same lengths. Did you draw a picture? That's step one in attacking word problems...
$endgroup$
– Eric Towers
Jan 27 '14 at 6:26
add a comment |
$begingroup$
how did you get 5x + 2y? Where does the 5 and the 2 come from? Also, how did you get 375 from 1500?
$endgroup$
– Jessica
Jan 26 '14 at 14:22
1
$begingroup$
@Jessica: Draw the picture of the $1 times 4$ division of the plot. Then count segments with the same lengths. Did you draw a picture? That's step one in attacking word problems...
$endgroup$
– Eric Towers
Jan 27 '14 at 6:26
$begingroup$
how did you get 5x + 2y? Where does the 5 and the 2 come from? Also, how did you get 375 from 1500?
$endgroup$
– Jessica
Jan 26 '14 at 14:22
$begingroup$
how did you get 5x + 2y? Where does the 5 and the 2 come from? Also, how did you get 375 from 1500?
$endgroup$
– Jessica
Jan 26 '14 at 14:22
1
1
$begingroup$
@Jessica: Draw the picture of the $1 times 4$ division of the plot. Then count segments with the same lengths. Did you draw a picture? That's step one in attacking word problems...
$endgroup$
– Eric Towers
Jan 27 '14 at 6:26
$begingroup$
@Jessica: Draw the picture of the $1 times 4$ division of the plot. Then count segments with the same lengths. Did you draw a picture? That's step one in attacking word problems...
$endgroup$
– Eric Towers
Jan 27 '14 at 6:26
add a comment |
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$begingroup$
What are 375 m and 600 m ? They don't appear anywhere.
$endgroup$
– Claude Leibovici
Jan 26 '14 at 8:14
$begingroup$
If $ x $ and $ y $ represent the dimensions of each plot, how many lengths of fence and how many widths are needed for the four parallel rectangles? (The answers are not 3 and 3...) Is there a picture that goes with this -- it matters whether the four smaller rectangles are parallel or arranged 2-by-2.
$endgroup$
– colormegone
Jan 26 '14 at 8:18