130 0 CHAPTER 16
16.53
we eventually obtain the equation (*) alx = bl(c - x), x = acl(a + b). To see that this yields the absolute
minimum, computation of /"(*) yields (after extensive simplifications) a
2 /(a
2 + x
2 )
3 '
2 + b
2 /[b
2 + (c- -t)
2 ]
3 '
2 ,
which is positive. [Notice that the equation (*) also tells us that the angles a and ft are equal. If we reinterpret
this problem in terms of a light ray from A being reflected off a mirror to B, we have found that the angle of
incidence a is equal to the angle of reflection 13.]
A telephone company has to run a line from a point A on one side of a river to another point B that is on the other
side, 5 miles down from the point opposite A (Fig. 16-21). The river is uniformly 12 miles wide. The
company can run the line along the shoreline to a point C and then run the line under the river to B. The cost of
laying the line along the shore is $1000 per mile, and the cost of laying it under water is twice as great. Where
should the point C be located to minimize the cost?
16.54
16.55
Fig. 16-21
I Let x be the distance from A to C. Then the cost of running the line is
The first-derivative test shows that this yields a relative maximum, and, therefore, by the uniqueness of the critical
number, an absolute maximum. When x = mS/(m + n), y = nSI(m + n).
Show that of all triangles with given base and given area, the one with the least perimeter is isosceles. (Compare
with Problem 16.20.)
I Let the base of length 2c lie on the jc-axis with the origin as its midpoint, and let the other vertex (x, y) lie in
the upper half-Diane (Fie. 16-22). By symmetry we may assume x £: 0. To minimize the perimeter, we must
±(c-x). The minus sign leads to the contradiction c = 0. Therefore, c + x = c-x, x = 0. Thus, the
third vertex lies on the y-axis and the triangle is isosceles. That the unique critical number x = 0 yields an
absolute minimum follows from a computation of the second derivative, which turns out to be positive.
Fig. 16-22
Setting" f'(x) = 0,
Hence,
Let x be the distance of S from C. Then the sum of the distances from A and B to S is given by the function
and
Setting
and solving
for x,
Since x cannot be negative or greater than 5, neither critical number is
feasible. So, the minimum occurs at an endpoint. Since /(O) = 26,000 and /(5) = 29,000, the minimum
occurs at x = 0.
Let m and n be given positive integers. If x and y are positive numbers such that x + y is a constant 5, find the
values of x and y that maximize P = x
m y".
Setting D V P = 0, we obtain x = mSI(m + n).
the altitude. Then
minimize AC + BC, which is given by the function
where h is
Setting
we obtain c -t- x =
f'(x)=0
f'(x)=0
P=x"'(s-x)". DxP=mx"'-1(S-x)"-nx"'(xS-x)n-1
16.53
we eventually obtain the equation (*) alx = bl(c - x), x = acl(a + b). To see that this yields the absolute
minimum, computation of /"(*) yields (after extensive simplifications) a
2 /(a
2 + x
2 )
3 '
2 + b
2 /[b
2 + (c- -t)
2 ]
3 '
2 ,
which is positive. [Notice that the equation (*) also tells us that the angles a and ft are equal. If we reinterpret
this problem in terms of a light ray from A being reflected off a mirror to B, we have found that the angle of
incidence a is equal to the angle of reflection 13.]
A telephone company has to run a line from a point A on one side of a river to another point B that is on the other
side, 5 miles down from the point opposite A (Fig. 16-21). The river is uniformly 12 miles wide. The
company can run the line along the shoreline to a point C and then run the line under the river to B. The cost of
laying the line along the shore is $1000 per mile, and the cost of laying it under water is twice as great. Where
should the point C be located to minimize the cost?
16.54
16.55
Fig. 16-21
I Let x be the distance from A to C. Then the cost of running the line is
The first-derivative test shows that this yields a relative maximum, and, therefore, by the uniqueness of the critical
number, an absolute maximum. When x = mS/(m + n), y = nSI(m + n).
Show that of all triangles with given base and given area, the one with the least perimeter is isosceles. (Compare
with Problem 16.20.)
I Let the base of length 2c lie on the jc-axis with the origin as its midpoint, and let the other vertex (x, y) lie in
the upper half-Diane (Fie. 16-22). By symmetry we may assume x £: 0. To minimize the perimeter, we must
±(c-x). The minus sign leads to the contradiction c = 0. Therefore, c + x = c-x, x = 0. Thus, the
third vertex lies on the y-axis and the triangle is isosceles. That the unique critical number x = 0 yields an
absolute minimum follows from a computation of the second derivative, which turns out to be positive.
Fig. 16-22
Setting" f'(x) = 0,
Hence,
Let x be the distance of S from C. Then the sum of the distances from A and B to S is given by the function
and
Setting
and solving
for x,
Since x cannot be negative or greater than 5, neither critical number is
feasible. So, the minimum occurs at an endpoint. Since /(O) = 26,000 and /(5) = 29,000, the minimum
occurs at x = 0.
Let m and n be given positive integers. If x and y are positive numbers such that x + y is a constant 5, find the
values of x and y that maximize P = x
m y".
Setting D V P = 0, we obtain x = mSI(m + n).
the altitude. Then
minimize AC + BC, which is given by the function
where h is
Setting
we obtain c -t- x =
f'(x)=0
f'(x)=0
P=x"'(s-x)". DxP=mx"'-1(S-x)"-nx"'(xS-x)n-1