448
Chapter 4 | Applications of Derivatives
1 2 4− x 2
A ′( x ) =2 4− x 2 +2 x ·
(−2 x )
=2 4− x 2 − 2 x 2
4− x 2
2 4− x 2
= 8−4 x
.
To find critical points, we need to find where A ′( x ) =0. We can see that if x is a solution of 8−4 x 2 4− x 2 =0, then x must satisfy 8−4 x 2 =0.
(4.7)
Therefore, x 2 =2. Thus, x =± 2 are the possible solutions of Equation 4.7 . Since we are considering x over the interval [0, 2], x = 2 is a possibility for a critical point, but x =− 2 is not. Therefore, we check whether 2 is a solution of Equation 4.7 . Since x = 2 is a solution of Equation 4.7 , we conclude that 2 is the only critical point of A ( x ) in the interval [0, 2]. Therefore, A ( x ) must have an absolute maximum at the critical point x = 2. To determine the dimensions of the rectangle, we need to find the length L and the width W . If x = 2 then
y = 1− ( 2) 2
1 2 =
1
.
4 = 1−
2
Therefore, the dimensions of the rectangle are L =2 x =2 2 and W =2 y = 2 2
= 2. The area of this
rectangle is A = LW =(2 2)( 2)=4.
4.35 Modify the area function A if the rectangle is to be inscribed in the unit circle x 2 + y 2 =1. What is the domain of consideration?
Solving Optimization Problems when the Interval Is Not Closed or Is Unbounded In the previous examples, we considered functions on closed, bounded domains. Consequently, by the extreme value theorem, we were guaranteed that the functions had absolute extrema. Let’s now consider functions for which the domain is neither closed nor bounded. Many functions still have at least one absolute extrema, even if the domain is not closed or the domain is unbounded. For example, the function f ( x ) = x 2 +4 over (−∞, ∞) has an absolute minimum of 4 at x =0. Therefore, we can still consider functions over unbounded domains or open intervals and determine whether they have any absolute extrema. In the next example, we try to minimize a function over an unbounded domain. We will see that, although the domain of consideration is (0, ∞), the function has an absolute minimum. In the following example, we look at constructing a box of least surface area with a prescribed volume. It is not difficult to show that for a closed-top box, by symmetry, among all boxes with a specified volume, a cube will have the smallest surface area. Consequently, we consider the modified problem of determining which open-topped box with a specified volume has the smallest surface area.
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