156 STEP 4. Review the Knowledge You Need to Score High
Step 3: Using the distance formula,Z=
√
(x− 19 )^2 +(y− 0 )^2 =√
(x− 19 )^2 +(x^2 − 2 x+ 1 − 0 )^2=
√
(x− 19 )^2 +(
(x− 1 )^2) 2
=√
(x− 19 )^2 +(x− 1 )^4.(Special case: In distance problems, the distance and the square of the distance
have the same maximum and minimum points.) Thus, to simplify computations,
letL=Z^2 =(x− 19 )^2 +(x− 1 )^4. The domain ofLis(−∞,∞).Step 4: Differentiate:
dL
dx
= 2 (x− 19 )( 1 )+ 4 (x− 1 )^3 ( 1 )
= 2 x− 38 + 4 x^3 − 12 x^2 + 12 x− 4 = 4 x^3 − 12 x^2 + 14 x− 42
=2(2x^3 − 6 x^2 + 7 x−21).
dL
dx
is defined for all real numbers.Set
dL
dx
=0; 2x^3 − 6 x^2 + 7 x− 21 =0. The factors of 21 are±1,±3,±7,
and ±21.
Using Synthetic Division, 2x^3 − 6 x^2 + 7 x− 21 =(x−3)(2x^2 +7)= 0 ⇒x=3.
Thus, the only critical number isx=3.
(Note: Step 4 could have been done using a graphing calculator.)
Step 5: Apply the First Derivative Test.03L′ – 0L+
[
decr incr
rel. minStep 6: Sincex=3 is the only relative minimum point in the interval, it is the absolute
minimum.Step 7: Atx=3,Z=√
( 3 − 19 )^2 +( 32 −2(3)+ 1 )^2 =√
(− 16 )^2 +( 4 )^2
=√
272 =√
16√
17 = 4√
17 .Thus, the shortest distance is 4√
17.TIP • Simplify numeric or algebraic expressions only if the question asks you to do so.Area and Volume Problem
Example Area Problem
The graph ofy=−1
2
x+2 encloses a region with thex-axis andy-axis in the first quadrant.
A rectangle in the enclosed region has a vertex at the origin and the opposite vertex on the
graph ofy=−1
2
x+2. Find the dimensions of the rectangle so that its area is a maximum.