SOLUTION: Randy is building a fence at the side of his warehouse. He has 120 m of fencing and plans to use the side of the warehouse as one side of the rectangular fenced area. What are the

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Randy is building a fence at the side of his warehouse. He has 120 m of fencing and plans to use the side of the warehouse as one side of the rectangular fenced area. What are the       Log On


   

Question 24591: Randy is building a fence at the side of his warehouse. He has 120 m of fencing and plans to use the side of the warehouse as one side of the rectangular fenced area. What are the dimensions of the maximum area Randy can enclose?
* I REALLY NEED HELP WITH THIS>.its due tomorrow
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Is this for a calculus class?
You don't need calculus to solve it anyway
Always draw a picture
|------------------|
|wwwwwwwwarew|
|wwwwwwhouseww|
|------------------|
|aaaafencedaaaaaa|
|------------------|
call the sides of the fenced area x. both sides are equal in a rectangle
the sum of the 2 sides is 2x
if there are 120m of fencing, the third side is 120 - 2x
so the equation for area is
A+=+x%2A%28120+-+2%2Ax%29
expanding
A+=+120+%2A+x+-+2%2Ax%5E2
try plugging in values for x to see what A is
x = 10
A+=+120+%2A+10+-+2+%2A+%2810%5E2%29
A = 1000
x = 20
A+=+120+%2A+20+-+2+%2A+%2820%5E2%29
A = 1600
x = 40
A+=+120+%2A+40+-+2+%2A+%2840%5E2%29
A = 1600
I get the same answer for x = 20 and x = 40, so the max must be in between,
since the curve is a parabola
x = 30
A+=+120+%2A+30+-+2+%2A+%2830%5E2%29
A = 1800
that is the max area in square meters, so the sides are 30 m and the third side
is 120 - 2(30) = 60 meters
check: 30 * 60 = 1800 m^2 for the area
and 2 * 30 + 60 = 120 for the length