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SOLUTIONS TO CHAPTER 12 MP REVIEW PROBLEMS
1. f''(x) = e
x
≥
0 for all x. Thus f(x) is a convex function.
2. Let x = location of store. We wish to choose x to minimize
f(x) = (x  3)
2
+ (x  4)
2
+ (x  5)
2
+ (x  6)
2
+ (x  17)
2
f'(x) = 2[ x  3 + x 4 + x  5 + x  6 + x  17] = 0 for
3 + 4
+ 5 + 6 + 17
x =  = 7. now f''(x) = 10>0
5
so x = 7 is a local minimum. Also f''(x)>0 implies that f(x) is convex so
x = 7 does indeed minimize f(x). In general the store should be located at
the arithmetic mean of the location of the n customers (i.e. x = (x
1
+ x
2
+ .
.. x
n
)/n).
3a. Let R = units of raw material purchased.
We wish to solve max z = (49  x
1
)x
1
+ (30  2x
2
)x
2
 5R
st (1) x
1
≤
2R
(2) x
2
≤
R,
All variables
≥
0
Since the objective function is concave and the constraints are linear the
KT conditions
will yield an optimal solution. The K T conditions are
(3) 49  2x
1

λ
1
= 0
(4) 30  4x
2

λ
2
= 0
(5) 5 + 2
λ
1
+
λ
2
= 0
(6)
λ
1
(x
1
 2R) = 0
(7)
λ
2
(x
2
 R) = 0
x
1
, x
2
,
λ
1
,
λ
2
≥
0
Let's try
λ
1
>0 and
λ
2
= 0. Then (6) yields x
1
= 2R. From (5)
λ
1
= 2.5. Then (3) yields x
1
= 23.25 and (4) yields x
2
= 7.5.
Since x
1
= 2R we find that R = 11.625. All KT conditions and original
constraints are satisfied so we have found an optimal solution. Since the
last unit of raw material purchased must have generated $5 in extra
revenue, an additional unit of raw material would be worth slightly less
than $5.
3b. R = 11.625, x
1
= 23.25, x
2
= 7.5.
4. For any two points on the curve y = x the line joining the two points
is never below the curve. Thus x is a convex function of x.
5. x
1
= 5  .618(5) = 1.91, x
2
= 0 + .618(5) = 3.09
f(x
1
) = 2.08>f(x
2
) = .28 so new interval of uncertainty is [0,
3.09). x
3
= 3.09  .618(3.09) = 1.18, x
4
= 1.91
f(x
3
) = 2.15 >f(x
4
) = 2.08 so new interval of uncertainty is
[0, 1.91). x
6
= 1.18, x
5
= 1.91  (.618)(1.91) = .73.
f(x
6
) = 2.15>f(x
5
) = 1.66 so new interval of uncertainty is
(.73, 1.91].
x
7
= 1.18 and x
8
= .73 + (.618)1.18 = 1.46.
1
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View Full Documentf(x
8
) = 2.25>f(x
7
) = 2.15 so new interval of uncertainty is
(1.18,
1.91]. Now x
9
= 1.46 and x
10
= 1.18 +.618(.73) = 1.63 and
f(x
9
) = 2.25>f(x
10
) = 2.23 so new interval of uncertainty is
[1.18, 1.63).
This interval has width less than .50, so we are finished. (actual maximum
occurs for x = 1.5)
6.
∇
f(x
1
, x
2
) =[exp(x
1
x
2
)(1 x
1
x
2
)1
exp(x
1
x
2
)(1 x
1
x
2
)].
∇
(0,1) = [1 0]. Thus new point is (t, 1) where t
≥
0
maximizes f(t)
= (1  t)e
t 1
+ t. f'(t) = (1  t)e
t1
 e
t1
+ 1
= 0 for 1 = te
t1
or t = 1.
thus new point is (1,1)
∇
f(1, 1) =
[0 1] so new point is [1
1+t] and
we choose t
≥
0 to maximize h(t) = te
t
+ 1, h'(t) = te
t
+ e
t
= 0 for t =
1. Thus new point is (1, 2).
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 Fall '07
 HOCHBAUM

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