This is one of the aplication questions I had trouble doing on a quadratics unit test. The question asks: A farmer wants to fence off three sides of a field adjacent to a farm wall. He has 220 of fencing. What dimensions should he use to maximize fenced off area. 

 

What I should have done is: 

1. Draw a diagram. We have two widths and one lenght since this is a three sided fence. The Information states 2w+l=220.

2. Write down an equation. So in this case, it's asking for the maximum possible area (optimization) so we would use A=l𝑥w. 

3. We want to isolate the variable l because it is going to make the problem a lot easier to solve. So now it would be l=220-2w.

4. So now plug l in the area equation and note that we only one varible other than A because we are trying to optimize the area. 

The equation looks like this: A=(220-2w)w

5. So now we are going to solve so we distribute w, which looks like this: A=220w-2w²

6. We now want the largest area possible and so we can look at the equation as a quadratic equation and graph it so that we can get the vertex and this will tell us the maximum value.

7. To get the vertex we use -b/2a so now our equation is:

w =-220/2(-2)

   =55

8. To get y, we plug it into 2(55)+l=220

                                         l=220-110

                                         l=110

Therefore, to get the maximum possible area we have to put 55m of fencing for width and 110m for lenght.