Question

A homeowner wants to fence a rectangular garden using 60ft. of fencing. An existing stone wall will be used as one side of the rectangle. Find the dimensions for which the area is a maximum.

Answer 1

since its a rectangle 2x+y= 60ft, as 1 side is a wall
the area of a rectangle is calculated by : xy
Isolates y from 2x+y = 60 we have y = 60-2x
Subsitute y = 60-2x for xy we have x(60-2x) = 60x-2x^2
Completing the square we (-2x^2+60x) -> -2(x^2-30x) -> -2(x-15)^2+450
Then the maximum point will be (15, 450)
Thus x = 15, y = 30 as 2x+y = 60
Maximum area is 450ft squqre

Answer 2

since its a rectangle 2x+y= 60ft, as 1 side is a wall
the area of a rectangle is calculated by : xy
Isolates y from 2x+y = 60 we have y = 60-2x

Now the Calculus way...

Set the Area as a function of x, and substitute what you got for y in the other equation
A(x)=x(60-2x)
A(x)=60x-2x^2
Now differentiate--
A'(x)=60-4x
And set A'(x)=0
0=60-4x
4x=60
x=15

To determine if it's the max, find the second derivative.
A"(x)=-4x
A"(15)=-4*15
A"15)=-60

Since the second derivative is negative, you know the function is concave down at that point, making 15 a maximum.

Now use the original information to find y.