126 0 CHAPTER 16
I Let the time / be measured in hours from 9 a.m. Choose a coordinate system with B moving along the jt-axis
and A moving along the _y-axis. Then the ^-coordinate of B is 65 - lOt, and the y-coordinate of B is -15f.
Let u be the distance between the ships. Then u
2 = (\5t)
2 + (65 - IQt)
2 . It suffices to minimize u
2 .
D,(u
2 ) = 2(150(15) + 2(65 - I0t)(-10) = 650f- 1300, and D,
2 («
2 ) = 650. Setting D,(w
2 ) = 0, we obtain
t = 2, Since the second derivative is positive, the unique critical number yields an absolute minimum. Hence,
the ships will be closest at 11 a.m.
Fig. 16-12
Fig. 16-13
16.39
A wall 8 feet high is 3.375 feet from a house. Find the shortest ladder that will reach from the ground to the
house when leaning over the wall.
I Let x be the distance from the foot of the ladder to the wall. Let y be the height above the ground of the point
where the ladder touches the house. Let L be the length of the ladder. Then L
2 = (x + 3.37S)
2 + y
2 .
It suffices to minimize L
2
By similar triangles, y/8 = (x + 3.375)Ix. Then D x y = -27/x
2 . Now,
D X (L
2 ) = 2(x + 3.375) + 2yD t y = 2(.v + 3.375) + 2(8/x)(x + 3.375)(-27Ix
2 ) = 2(x + 3.375)(1 - 216/*
3 ). Solving D r (L
2 ) = 0, we find the unique positive critical number x = 6. Calculation of £>
2 (L
2 ) yields 2 +
(
2 /*
4 )[(27)
2 + yx] > 0. Hence, the unique positive critical number yields the minimum length. When x = 6,
y=f, L =-^ = 15.625 ft.
Fig. 16-14
16.40
A company offers the following schedule of charges: $30 per thousand for orders of 50,000 or less, with the
charge per thousand decreased by 37.5 cents for each thousand above 50,000. Find the order that will maximize
the company's income.
I Let x be the number of orders in thousands. Then the price per thousand is 30 for x s 50 and
30-j|(jt-50) for *>50. Hence, for *<50, the income 7 = 30*, and, for jt>50, / = ;t[30g(;t-50)]= ™x - Ix
2 . So, for x<50, the maximum income is 1500 thousand. For *>50, DJ =
ir ~ !* and D
2 /=—|. Solving D X I = 0, A: = 65. Since the second derivative is negative, x = 65
yields the maximum income for x > 50. That maximum is 3084.375 thousand. Hence, the maximum income
is achieved when 65,000 orders are received.
25 + x , 3x
2 = 25, x = 5V3/3 = 2.89. Since the second derivative is positive, the unique critical number yields
the absolute minimum time.
we obtain
and the
distance walked is
Hence the total time
Let x be the distance between A and the landing point. Then the distance rowed is
16.38
A woman in a rowboat at P, 5 miles from the nearest point A on a straight shore, wishes to reach a point B, 6 miles
from A along the shore (Fig. 16-13). If she wishes to reach B in the shortest time, where should she land if she
can row 2 mi/h and walk 4 mi/h?
Then
Setting
I Let the time / be measured in hours from 9 a.m. Choose a coordinate system with B moving along the jt-axis
and A moving along the _y-axis. Then the ^-coordinate of B is 65 - lOt, and the y-coordinate of B is -15f.
Let u be the distance between the ships. Then u
2 = (\5t)
2 + (65 - IQt)
2 . It suffices to minimize u
2 .
D,(u
2 ) = 2(150(15) + 2(65 - I0t)(-10) = 650f- 1300, and D,
2 («
2 ) = 650. Setting D,(w
2 ) = 0, we obtain
t = 2, Since the second derivative is positive, the unique critical number yields an absolute minimum. Hence,
the ships will be closest at 11 a.m.
Fig. 16-12
Fig. 16-13
16.39
A wall 8 feet high is 3.375 feet from a house. Find the shortest ladder that will reach from the ground to the
house when leaning over the wall.
I Let x be the distance from the foot of the ladder to the wall. Let y be the height above the ground of the point
where the ladder touches the house. Let L be the length of the ladder. Then L
2 = (x + 3.37S)
2 + y
2 .
It suffices to minimize L
2
By similar triangles, y/8 = (x + 3.375)Ix. Then D x y = -27/x
2 . Now,
D X (L
2 ) = 2(x + 3.375) + 2yD t y = 2(.v + 3.375) + 2(8/x)(x + 3.375)(-27Ix
2 ) = 2(x + 3.375)(1 - 216/*
3 ). Solving D r (L
2 ) = 0, we find the unique positive critical number x = 6. Calculation of £>
2 (L
2 ) yields 2 +
(
2 /*
4 )[(27)
2 + yx] > 0. Hence, the unique positive critical number yields the minimum length. When x = 6,
y=f, L =-^ = 15.625 ft.
Fig. 16-14
16.40
A company offers the following schedule of charges: $30 per thousand for orders of 50,000 or less, with the
charge per thousand decreased by 37.5 cents for each thousand above 50,000. Find the order that will maximize
the company's income.
I Let x be the number of orders in thousands. Then the price per thousand is 30 for x s 50 and
30-j|(jt-50) for *>50. Hence, for *<50, the income 7 = 30*, and, for jt>50, / = ;t[30g(;t-50)]= ™x - Ix
2 . So, for x<50, the maximum income is 1500 thousand. For *>50, DJ =
ir ~ !* and D
2 /=—|. Solving D X I = 0, A: = 65. Since the second derivative is negative, x = 65
yields the maximum income for x > 50. That maximum is 3084.375 thousand. Hence, the maximum income
is achieved when 65,000 orders are received.
25 + x , 3x
2 = 25, x = 5V3/3 = 2.89. Since the second derivative is positive, the unique critical number yields
the absolute minimum time.
we obtain
and the
distance walked is
Hence the total time
Let x be the distance between A and the landing point. Then the distance rowed is
16.38
A woman in a rowboat at P, 5 miles from the nearest point A on a straight shore, wishes to reach a point B, 6 miles
from A along the shore (Fig. 16-13). If she wishes to reach B in the shortest time, where should she land if she
can row 2 mi/h and walk 4 mi/h?
Then
Setting