Math 241
Lab Exercise: Smokestack Problem
Problem: A smokestack deposits soot on the ground with a concentration inversely proportional to the square of the
distance from the stack. With two smokestacks d miles apart, the concentration of the combined deposits on the line
joining them, at a distance x from one stack, is given by
S(x) =
c
k
+
2
x
(d − x)2
where c and k are positive constants which depend on the quantity of smoke each stack is emitting. If k = 8c, find the
point on the line joining the stacks where the concentration of the deposit is a minimum.
Solution:
We are given a function to minimize:
S(x) = cx−2 + k(d − x)−2
and we’re told that k = 8c, so
S(x) = c[x−2 + 8(d − x)−2 ]
where we’ve factored out the c. Now, we are also told that c is a positive number, so we can leave it out of the rest of
the problem without effecting the answer, if we want (but we could leave it in there and still get the right answer). So
here’s the function we will minimize:
s(x) = x−2 + 8(d − x)−2
The derivative is
s0 (x) = −2x−3 + 8(−2)(d − x)−3 (−1)
(Don’t forget the last −1 which accounts for the derivative of the “inside function” d − x in the chain rule.) We set
s0 (x) equal 0 and solve for x as follows:
2x−3 = 16(d − x)−3
Equivalently,
(d − x)3
16
=
,
x3
2
or,
(d − x)3
= 8.
x3
Now take the cube root of both sides to get
(d − x)
= 81/3 = 2.
x
Equivalently, d − x = 2x, or 3x = d. So the critical number is x = d/3.
Important Point
Notice that all we have shown so far is that S(d/3) is a relative extremum. We must show that it is an absolute
extremum, and also we must show that it is, indeed, a minimum (and not a maximum) value of S(x). We check that
S(d/3) is an absolute extremum by comparing it to the value of S(x) at other critical numbers in the domain, and
at the endpoints of the domain (or by taking the limit of S(x) as x approaches the endpoints of the domain). In this
problem, x = d/3 is the only critical number, so we need only compare the value S(d/3) with the value of S(x) as x
approaches the endpoints of the domain.
Finally, check that S(d/3) is a minimum (and not a maximum) by looking at the second derivative of s(x):
s00 (x) = 6x−4 + 48(d − x)−4 .
If s00 (x) > 0 at the point x = d/3, the function is concave up and we have found a minimum, or s00 (d/3) < 0, in
which case the function is concave down and we have found a maximum.
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