Part II: The Answers
326
Answers
501–600
Use x = 20 4
3
to find y:
y =
=
=
=
= ( )
16 000
20 4
16 000
400 4
40
4
40 4
4
10 4
1 3 2
2 3
2 3
1 3
1 3
,
( )
,
(
)
( )
(
)
( )
= = 10 4
3
Therefore, the dimensions are 20 4
3
cm × 20 4
3
cm × 10 4
3
cm (approximately 31.75 cm ×
31.75 cm × 15.87 cm).
Note that you can easily use the second derivative test to verify that x = 20 4
3
gives
you a minimum because S" > 0 for all x > 0.
504.
− 1
7
8 6
7
,

 

  , − −
1
7
8 6
7
,

 

 
You want to find the maximum distance. The distance from a point (x, y) to the point
(1, 0) is given by
D x y
x
y
x
x
y
( , )
(
)
=
−
+ ( − )
=
−
+ +
1
0
2
1
2
2
2
2
Rewrite the equation of the ellipse as y
2
= 8 – 8x
2
and substitute the value of y
2
into the
distance equation:
D x
x
x
x
x
x
( ) =
−
+ + −
=
−
−
2
2
2
2 1 8 8
9 2
7
Tip: You can take the derivative of this function and use the first derivative test to find
a maximum, but it’s easier to use the square of the distance, which gets rid of the radical. For a function that satisfies f
 
(x) ≥ 0, its local maxima and minima occur at the same
x values as the local maxima and minima of its square, [f
 
(x)]
2
. Obviously, the corresponding y values would change, but that doesn’t matter here!
The square of the distance is
S
x
x
= −
−
9 2
7
2
And the derivative of this function is
′
S
x
= − −
2 14
Setting the derivative equal to zero and solving gives you –2 – 14x = 0, or –2 = 14x,
which has the solution x = − 1
7
. You can verify that x = − 1
7
gives you a maximum by
using the first derivative test.