SOLUTION: I need help with setting up the function. Thanks. A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 328 feet of fe

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: I need help with setting up the function. Thanks. A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 328 feet of fe      Log On


   



Question 668721: I need help with setting up the function. Thanks.
A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 328 feet of fencing and doesn’t fence the side along the street, what is the largest area that can be enclosed?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
This is like fencing in a rectangle, then removing one side
Let +A+ = the area enclosed
Let +y+ = the side parallel to the street
Let +x+ = the length of a side perpendicular to the street
The amount of fencing is +y+%2B+2x+
(1) +y+%2B+2x+=+328+
The area enclosed is
(2) +A+=+x%2Ay+
----------------
(1) +y+=+328+-+2x+
By substitution:
(2) +A+=+x%2A%28+328+-+2x+%29+
(2) +A+=+-2x%5E2+%2B+328x+
Because of the minus sign in front of the squared term, the equation
has a maximum and not a minimum
(2) +A+=+x%2A%28+-2x+%2B+328+%29+
The maximum occurs halfway between the 2 roots
One root is at +x+=+0+
and to find the other root,
+-2x+%2B+328+=+0+
+2x+=+328+
+x+=+164+
Half way between is +x+=+82+
and, +y+=+328+-+2x+
+y+=+328+-+2%2A82+
+y+=+328+-+164+
+y+=+164+
and, at the maximum,
(2) +A+=+x%2Ay+
(2) +A+=+82%2A164+
(2) +A+=+13448+ ft2
---------------------
Here's the plot with +A+ on the vertical axis
and +x+ on the horizontal axis
+graph%28+400%2C+400%2C+-50%2C+200%2C+-3000%2C+15000%2C+-2x%5E2+%2B+328x+%29+