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**I made up this problen and I want to see if you can solve it...please help**

 

Suppose you want to enclose a rectangular garden plot against a house using fencing on three sides. Assume you have 50 ft of fencing material and want to create a garden with an area of 150ft². 

  • a. Let w = the width. Write an expression for th length of the plot 
  • b. Write and solve an equation for the area of the plot. Round to the nearest tenth of a foot. 
  • c. What dimensions should the garden have? 
  • d. Find the area of the garden by using the dimensions you found in part (b). Does the area equal 150 ft²? 
  • e. Explain.
 Mar 22, 2016

Best Answer 

 #1
avatar+2592 
+5

Ummm how about Length(l) = 4/3w

Then it is just width =15 and then length 20. With the length parallel to the house.

 Mar 22, 2016
 #1
avatar+2592 
+5
Best Answer

Ummm how about Length(l) = 4/3w

Then it is just width =15 and then length 20. With the length parallel to the house.

SpawnofAngel Mar 22, 2016
 #2
avatar+129849 
+5

We have that a perimeter that is composed of a width and two congruent lengths of fencing

 

So we have

 

50 = w + 2l     subtract w from both sides

 

50 - w = 2l     divide both sides by 2

 

[ 50 - w] / 2   = l

 

 

And area  = l * w    so we have

 

150  = [50 - w] / 2   * w   multiply both sides by 2

 

300  = [ 50 - w ] * w      simplify

 

300 = 50w - w^2      rearrange

 

w^2 - 50w + 300 = 0   and using the quadratic formula....

 

w  = 25 + 5√13 ft   [about 43.03 ft]

 

And the length  = [50 - (25 + 5√13) ] / 2    ft   =   

 

  [ 25 - 5√13] / 2    ft    =   [about 3.487 ft]

 

Check the area = l * w  = 150 ft^2  ???

 

[ 25 + 5√13] * [ 25 - 5√13] / 2   =   150 ft^2

 

 

cool cool cool

 Mar 22, 2016

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