Word problem involving quadratic?

One Response to Word problem involving quadratic?

1. Gary B | September 3, 2010 at 3:30 pm |

   So, do you mean he uses the fence for 3 sides of the lot?
   So, the 280 meters will equal the 3 sides of the perimeter
   2w+l=280
   Solve for l, so we have an equation with only one variable
   l=280-2w
   Now, the area is l*w=A
   Substitute the value of l into the area equation
   A=w*(280-2w)
   A=280w-2w^2
   A=-2w^2+280w
   We now have a quadratic equation. The parabola of this equation will open downward, with the highest point on the parabola being the vertex. We want to find this vertex because it represents the maximum area. The x value will represent what value of w will give the maximum area, and the y value will represent the maximum area itself
   To find the vertex, we simply need to find (-b/2a,A(-b/2a)) for the equation in form ax^2+bx+c
   So, the ‘b’ in the equation is 280. The ‘a’ in the equation is -2
   -280/2(-2)=-280/-4=70
   So, we will want a width of 70 meters for the lot. To find the maximum area itself, reinsert this value into the equation A=-2w^2+280w
   A=-2(70)^2+280(70)
   A=-2(4900)+280(70)
   A=-9800+19600
   A=9800. So, the maximum area is 9,800 square meters for the lot

   The dimensions:
   well, we know we want the width to be 70.
   Our perimeter equation was 280=l+2w
   280=l+2(70)
   280=l+140
   l=140. So, the width will be 70 meters, and the length will be 140 meters
   The amount of barn wall used will be 140 meters for the other length

   Let’s check. You have 280 meters of fence. You will use 70 m on each width, and 140 m on the length:
   70+70+140=280. So, it checks

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