Question Regarding System of Equations

In summary, the problem involves two rectangular fields enclosed by 600m of fencing, with one field having fencing on only three sides due to a river boundary. The goal is to create a system of quadratic equations to model the problem, with the answers being A = -x2+300x and A = -2x2+60x. The incorrect equations and solutions provided by the student were due to the use of the same variables for both fields. Using different variables for each field yields the correct equations and solutions.
  • #1
trulyfalse
35
0
Hey Pf, I'm working on precalculus review and I have found myself stumped by a question.

Homework Statement



A rectangular field is enclosed by 600m of fencing. A second rectangular field, which is alongside a river, has the same area and is also enclosed by 600m of fencing. However, this second field has fencing on only three sides because there is no need for fencing along the riverbank. Create a system of quadratic equations to model the problem. (Answer: A = -x2+300x, A = -2x2+60x)

Homework Equations



ax2+bx+c

The Attempt at a Solution



I began by modeling the first field:
P = 2x+2y
600 = 2x+2y
300 = x+y
y = -x+300

A = xy
A = x(-x+300)
A = -x2+300x

Afterwards I modeled the second field:
600 = 2x+y
y = -2x+600

A = xy
A = x(-2x+600)
A = -2x2+600x

As you can see, the equations are different. I cannot see where I went wrong. Could someone please correct my folly? :)
 
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  • #2
I can't see where you wrong, are you sure you read the answer right?
 
  • #3
Positive. It may be of use to note when I entered the equations I created into my calculator it yielded solutions of (0,0) and (0,300). Seeing as those solutions are illogical, that means it's either a bad question, or I made an error somewhere along the way.

EDIT: I just used the same method of checking for the "correct" solution; it yielded values of (-240,-129600) and (0,0). Perhaps it's just a bad question...
 
  • #4
Well the area of the second one should be [itex]\geq[/itex] the first one. And if the answers are right, that doesn't always hold true.
 
  • #5
I'll skip it and move on I suppose. :)
 
  • #6
Yeah that might be a good idea.
 
  • #7
trulyfalse said:
Hey Pf, I'm working on precalculus review and I have found myself stumped by a question.

Homework Statement



A rectangular field is enclosed by 600m of fencing. A second rectangular field, which is alongside a river, has the same area and is also enclosed by 600m of fencing. However, this second field has fencing on only three sides because there is no need for fencing along the riverbank. Create a system of quadratic equations to model the problem. (Answer: A = -x2+300x, A = -2x2+60x)

Homework Equations



ax2+bx+c

The Attempt at a Solution



I began by modeling the first field:
P = 2x+2y
600 = 2x+2y
300 = x+y
y = -x+300

A = xy
A = x(-x+300)
A = -x2+300x

Afterwards I modeled the second field:
600 = 2x+y
y = -2x+600

A = xy
A = x(-2x+600)
A = -2x2+600x

As you can see, the equations are different. I cannot see where I went wrong. Could someone please correct my folly? :)

You can't use the same variables for the two fields.
Let x1 and y1 be the width and length of the first field (the one with fences along all four sides).
Let x2 and y2 be the width and length of the second field (the one with the river as one boundary).

For the first field, A = x1*y1 = x1(300 - x1)
For the second field, A = x2*y2 = x2(600 - 2x2)
 

1. How do you solve a system of equations?

To solve a system of equations, you need to find values for each variable that satisfy both equations. This can be done by elimination, substitution, or graphing.

2. What is the difference between a linear and a nonlinear system of equations?

A linear system of equations has only linear equations, meaning the variables are raised to the first power and there are no exponents. A nonlinear system of equations has at least one nonlinear equation, meaning the variables may be raised to a power other than 1 or may have exponents.

3. Can a system of equations have no solution?

Yes, a system of equations can have no solution if the equations are parallel lines or if they intersect at a point that does not satisfy both equations. This means that there is no set of values for the variables that will make both equations true at the same time.

4. How many solutions can a system of equations have?

A system of equations can have one solution, no solution, or infinitely many solutions. The number of solutions depends on the equations and their relationship to each other.

5. What is a consistent and inconsistent system of equations?

A consistent system of equations has at least one solution, meaning the equations intersect at one point. An inconsistent system of equations has no solutions, meaning the equations do not intersect or are parallel.

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