Math help with quadratic word problems

Question

A rancher uses 200m of fencing to enclose two adjacent rectangular corrals, forming one large corral with a fence in the interior. Determine the dimensions that enclose a total area of 1400m

Answer 1

 

           x

______________

|                       |

|                       |y

|                       |

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If I read this correctly we are merely finding the dimensions of

a rectangle with area 1400m² and perimeter of 200m

 

xy = 1400            {equation 1, area}

2x + 2y = 200      {equation 2, perimeter}

 

From equation 2:

2(x+y) = 200

x + y = 100

y = 100 - x

Substitute back into equation 1

 

x(100-x) = 1400

100x - x² = 1400

Rearrange into  quadratic of form ax² + bx + c = 0

 

x² - 100x + 1400 = 0

 

from quadratic equation x = [-b ±√(b²-4ac)]/2a  a=1, b=-100, c = 1400

 

x = [100  ±√((-100)²-4(1)(1400))]/2

x = [100 ±√(10000 - 5600)]/2

x = (100 ±√4400)/2

x = (100 ± 66.3325)/2

x = 83.16625  or   16.83375

y = 16.83375  or   83.16625

 

Rounding to one decimal, the corral is basically 83.2 m by 16.8 m

 

 

Answer 2

I made a drawing with the "width" being the dimension with three pieces of fence, and the "length" being the two other sides.

 

The total area is length X width.

 

The amount of fencing used is 3 widths and 2 lengths.  (Make sure you drawing reflects this - That will help it make sense.)

 

Now develop equations:

 

Area:

 

(L)(w) = 1400

 

2L + 3w = 200

 

 

Solve the first equation for L:

 

L = 1400/w

 

Now plug this into the second equation:

 

2(1400/w) + 3w = 200

 

Multiply all terms by "w":

 

2800 + 3w² = 200w

 

Rearrange into standard form:

 

3w² -200w + 2800 = 0

 

 

Use the quadratic formula to solve for w:

 

You get two answers:  w = 46.666667 and w = 20

 

 

Plug each solution for w into the first equation (L = 1400/w):

 

L = 30 and L = 70

 

 

So there are two different solutions to the problem.