MA 3972-MA-Book April 11, 2018 15:57
Def inite Integrals 269
x = – 3 x = – 1
y = 1
1
2
y = – 1
x
y
Figure 12.8-3
- (Calculator)
Step 1: (See Figure 12.8-4.) LetP=x+y
wherePis the length of the pipe
andxandyare as shown. The
minimum value ofPis the
maximum length of the pipe to be
able to turn in the corner. By
similar triangles,
y
10
=
x
√
x^2 − 36
and thus,y=
10 x
√
x^2 − 36
, x> 6
P=x+y=x+
10 x
√
x^2 − 36
.
x
y
10
6
x^2 – 36
Figure 12.8-4
Step 2: Find the minimum value ofP.
Enter
yt=x+ 10 ∗x/
(√
(x∧ 2 − 36 )
)
.
Use the [Minimum] function of
the calculator and obtain the
minimum point (9.306, 22.388).
Step 3: Verify with the First Derivative
Test.
Entery 2 =(y1(x),x) and
observe. (See Figure 12.8-5.)
9.306
rel. min.
decr. incr.
y 1 = f
y 2 = f′
Figure 12.8-5
Step 4: Check the endpoints.
The domain ofxis (6,∞).
Sincex= 9 .306 is the only
relative extremum, it is the
absolute minimum.
Thus, the maximum length of the
pipe is 22.388 feet.
