SOLUTION: 500 feet of fencing is available to enclose a rectangular lot along side of highway 65. Cal trans will supply the fencing for the side along the highway, so only 3 sides are needed

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 500 feet of fencing is available to enclose a rectangular lot along side of highway 65. Cal trans will supply the fencing for the side along the highway, so only 3 sides are needed      Log On


   



Question 1150524: 500 feet of fencing is available to enclose a rectangular lot along side of highway 65. Cal trans will supply the fencing for the side along the highway, so only 3 sides are needed. What dimensions will produce an area of 40,000 square feet? What is the maximum area that can be enclosed?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the side perpendicular to the highway = +w+
+500+-+2w+ = the length of the side parallel to the highway
Let +A+ = the area of the lot
-----------------------------------
+A+=+w%2A%28+500+-+2w+%29+
+A+=+-2w%5E2+%2B+500w+
Use the formula for the vertex:
+w%5Bpk%5D+=+-500%2F%28+2%2A%28-2%29%29+
+w%5Bpk%5D+=+125+
Plug this result back into equation
+A%5Bmax%5D+=+-2%2A%28125%29%5E2+%2B+500%2A125+
+A%5Bmax%5D+=+31250+ ft2
Since 40000 ft2 is greater than the maximum area
this can't be enclosed with 500 ft of fencing
Here's the plot:
+graph%28+400%2C+400%2C+-100%2C+300%2C+-3500%2C+35000%2C+-2x%5E2+%2B+500x+%29+