Part II: The Answers
334
Answers
501–600
You want to maximize the area of the rectangle. From the diagram, the area of the
rectangle is
A x y
x y
xy
( , ) = ( )( ) =
2 2
4
Using the equation of the ellipse, solve for x:
x
y
x
y
x
y
x
y
2
2
2
2
2
2
2
4
9
1
4
1 9
4
4
9
4
4
9
+
=
= −
= −
= ± −
Keeping only the positive solution, you have x
y
=
−
4
4
9
2
(although you can just as
easily work with the negative solution). Therefore, the area in terms of y becomes
A y
y
y
y
y
( )
( )
=
−
=
−
4 4 4
9
4 4 4
9
2
1 2
2
1 2
( )
( )
Take the derivative by using the product rule and the chain rule:
′
( )
( ) ( )






( )
A
y
y
y
y
y
y
=
−
+
−
−
=
−
−
−
4 4 4
9
4 1
2
4 4
9
8
9
4 4 4
9
16
2
1 2
2
1 2
2
1 2
2 2
2
1 2
2
2
2
1 2
2
2
9 4 4
9
36 4 4
9
16
9 4 4
9
144 32
9 4 4
9
−
=
−
−
−
=
−
−
y
y
y
y
y
y
(
)
(
)
(
)
(
)
1 1 2
Then set the derivative equal to zero and solve for y:
144 32
0
144 32
144
32
9
2
2
2
2
−
=
=
=
±
=
y
y
y
y