QR Word Problem - Capture Area

tom_servo_dds

Senior Member
10+ Year Member
Could someone please explain the Algebra approach to this problem:

A 120 m trapping net stretched out into a 3-sided rectangular corral is used to front a flowing river. What value of widths (w), being parallel to the riverbanks, will provide the corral with largest capturing area?

A. w = 15 m
B. w = 20 m
C. w = 25 m
D. w = 30 m
E. w = 35 m

I'm stuck as to how the author came up with an answer for this one. Thanks in advance!

DMD to Be

not luffin' this weather.
10+ Year Member
5+ Year Member
tom_servo_dds said:
Could someone please explain the Algebra approach to this problem:

A 120 m trapping net stretched out into a 3-sided rectangular corral is used to front a flowing river. What value of widths (w), being parallel to the riverbanks, will provide the corral with largest capturing area?

A. w = 15 m
B. w = 20 m
C. w = 25 m
D. w = 30 m
E. w = 35 m

I'm stuck as to how the author came up with an answer for this one. Thanks in advance!
I don't even understand what the question is asking, much less how to come up with an equation?

OP

tom_servo_dds

Senior Member
10+ Year Member
Anyone wanna take a crack at this one?

J

jkh1886

tom_servo_dds said:
Could someone please explain the Algebra approach to this problem:

A 120 m trapping net stretched out into a 3-sided rectangular corral is used to front a flowing river. What value of widths (w), being parallel to the riverbanks, will provide the corral with largest capturing area?

A. w = 15 m
B. w = 20 m
C. w = 25 m
D. w = 30 m
E. w = 35 m

I'm stuck as to how the author came up with an answer for this one. Thanks in advance!
The answer is C. You know why? ....it has a 22% chance of being correct in recent DAT's. Screw this question and move on!

LNinlove

Member
7+ Year Member
15+ Year Member
tom_servo_dds said:
Anyone wanna take a crack at this one?
Hah, wish I could help...it would be nice if they made the question a little clearer! Good luck...

djeffreyt

Senior Member
10+ Year Member
7+ Year Member
answer is explained correctly and done incorrectly.

The question is basically saying what shape rectangle given a certain amount of rope/net will have the greatest area. the answer to that is...a square. A sqaure using 120m length at max would be 30m on each side. however, the question is about capturing fish, so one side of the net must be open to catch the fish in, so the net length only goes along 3 sides of the square.

The answer should be 40 m for each side, and the explanation in the practice test gives the explanation correctly, but then it chooses 30m (the side if it was a 4 sided net) as the correct choise.

This is my 2 cents, but I could be wrong....in which case...guess

OP

tom_servo_dds

Senior Member
10+ Year Member
djeffreyt said:
answer is explained correctly and done incorrectly.

The question is basically saying what shape rectangle given a certain amount of rope/net will have the greatest area. the answer to that is...a square. A sqaure using 120m length at max would be 30m on each side. however, the question is about capturing fish, so one side of the net must be open to catch the fish in, so the net length only goes along 3 sides of the square.

The answer should be 40 m for each side, and the explanation in the practice test gives the explanation correctly, but then it chooses 30m (the side if it was a 4 sided net) as the correct choise.

This is my 2 cents, but I could be wrong....in which case...guess
It's not a very clear question, but thanks for taking a crack at it anyway!

Indie

Member
10+ Year Member
djeffreyt said:
answer is explained correctly and done incorrectly.

The question is basically saying what shape rectangle given a certain amount of rope/net will have the greatest area. the answer to that is...a square. A sqaure using 120m length at max would be 30m on each side. however, the question is about capturing fish, so one side of the net must be open to catch the fish in, so the net length only goes along 3 sides of the square.

The answer should be 40 m for each side, and the explanation in the practice test gives the explanation correctly, but then it chooses 30m (the side if it was a 4 sided net) as the correct choise.

This is my 2 cents, but I could be wrong....in which case...guess

I got A length of 30 W being the size that gives the maximum capture area... Im pretty sure im right, but before I make a mathematical idiot of myself.. Does anyone know what the correct answer is??

REH

Junior Member
10+ Year Member
5+ Year Member
My view on this problem is a little different...I could be waaaaay off, but here's how I look at it. We have a net of a fixed length of 120m. This net is going to span across river to create a three-sided rectangle. There are going to be two horizontal portions that run parallel to the river bank and one perpendicular portion to the river bank. This creates a three sides of a rectangle. It's the perpendicular portion, the capturing area, that we are trying to maximize. In essence, I think this question looks at the relationship between width and length. If we have a defined perimeter and want to maximize length, then there must be a proportional reduction in width. The length of this 3-sided rectangle would be maximized when width is the smallest, thus I'd choose 15m. That's my take...

REH

Indie

Member
10+ Year Member
REH said:
My view on this problem is a little different...I could be waaaaay off, but here's how I look at it. We have a net of a fixed length of 120m. This net is going to span across river to create a three-sided rectangle. There are going to be two horizontal portions that run parallel to the river bank and one perpendicular portion to the river bank. This creates a three sides of a rectangle. It's the perpendicular portion, the capturing area, that we are trying to maximize. In essence, I think this question looks at the relationship between width and length. If we have a defined perimeter and want to maximize length, then there must be a proportional reduction in width. The length of this 3-sided rectangle would be maximized when width is the smallest, thus I'd choose 15m. That's my take...

REH

I see your thinking here, but I think that would only work if the river was as wide as the net, and the fish were all swimming in the same direction....

REH

Junior Member
10+ Year Member
5+ Year Member
I guess when I assume I make an ass out of -u and -me

OP

tom_servo_dds

Senior Member
10+ Year Member
Indie said:
I got A length of 30 W being the size that gives the maximum capture area... Im pretty sure im right, but before I make a mathematical idiot of myself.. Does anyone know what the correct answer is??
I appologize for not posting it earlier, but the answer is 30W (D). Apparently there is a calculus method to getting this answer as well as an algebra one. I've never taken calculus, but the algebra answer is explained as:

Note that this is a 3-sided rectangular corral. {Two widths and one length}

Length of corral, l = 120  2w

Capturing area, A = l w = (120  2w)w = 120w  2w2

I. Using calculus approach:

120  4w = 0 => w = 30 m

II. The algebra way:

A = -2(w2 - 60w) = -2(w2 - 60w + 302 -302) = -2(w - 30)2 + 1800

Since A is less than or equal to 1800, it will max out when:

(w - 30) = 0 => w = 30 m

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luder98

Senior Member
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5+ Year Member
tom_servo_dds said:
I appologize for not posting it earlier, but the answer is 30W (D). Apparently there is a calculus method to getting this answer as well as an algebra one. I've never taken calculus, but the algebra answer is explained as:

Note that this is a 3-sided rectangular corral. {Two widths and one length}

Length of corral, l = 120  2w

Capturing area, A = l w = (120  2w)w = 120w  2w2

I. Using calculus approach:

120  4w = 0 => w = 30 m

II. The algebra way:

A = -2(w2 - 60w) = -2(w2 - 60w + 302 -302) = -2(w - 30)2 + 1800

Since A is less than or equal to 1800, it will max out when:

(w - 30) = 0 => w = 30 m
I saw this exact question on my DAT. If I'm not mistaken, this is from DAT achiever, right? Here is another algebra way to solve this problem.
A = (120  2w)w = 2(60-w)w

Notice: 60 - w + w = 60: constant
If the sum of two non-negative variables is constant, the product will be at max when the two are equal. Thus,
A is at max when 60 - w = w or 60 = 2w or w = 30.

PS: If the product is constant, the sum will be at min if the two are equal.

Indie

Member
10+ Year Member
tom_servo_dds said:
I appologize for not posting it earlier, but the answer is 30W (D). Apparently there is a calculus method to getting this answer as well as an algebra one. I've never taken calculus, but the algebra answer is explained as:

Note that this is a 3-sided rectangular corral. {Two widths and one length}

Length of corral, l = 120  2w

Capturing area, A = l w = (120  2w)w = 120w  2w2

I. Using calculus approach:

120  4w = 0 => w = 30 m

II. The algebra way:

A = -2(w2 - 60w) = -2(w2 - 60w + 302 -302) = -2(w - 30)2 + 1800

Since A is less than or equal to 1800, it will max out when:

(w - 30) = 0 => w = 30 m

Wow, good thing I kept my trap shut... I dont know what you guys are talking about.. I got the answer a much simpler way, but after seeing your guys explanations, it must have been coincidence.. Lets hope I get the same coincidence on my DAT.... Its good enough for me..

OP

tom_servo_dds

Senior Member
10+ Year Member
luder98 said:
I saw this exact question on my DAT. If I'm not mistaken, this is from DAT achiever, right? Here is another algebra way to solve this problem.
A = (120  2w)w = 2(60-w)w

Notice: 60 - w + w = 60: constant
If the sum of two variables is constant, the product will be at max when the two are equal. Thus,
A is at max when 60 - w = w or 60 = 2w or w = 30.

PS: If the product is constant, the sum will be at min if the two are equal.
Bless you luder98, bless you!