Download Homework from Section 4.5 4.5.3. Find two positive numbers whose

Homework from Section 4.5
4.5.3. Find two positive numbers whose product is 100 and whose sum is a minimum.
We want x and y so that xy = 100 and S = x + y is minimized. Since xy = 100, x 6= 0.
Thus we have y = 100/x, so we want to minimize S( x ) = x + 100/x. We find that
S0 ( x ) = 1 − 100/x2 . Setting S0 ( x ) = 0 and solving for x yields x = ±10. Since we are
looking for positive numbers, we take x = 10. Notice that for 0 < x < 10, x2 < 100, so
100/x2 > 1, which implies that 1 − 100/x2 < 0. Thus S0 ( x ) < 0 for such x. Similarly, for
x > 10, S0 ( x ) > 0. It follows from the first derivative test that x = 0 is a minimum of
S( x ). Thus x = 10 and y = 100/x = 10 is a pair of positive numbers whose product is
100 and whose sum is a minimum.
4.5.7. Consider the following problem: A farmer with 750 ft of fencing wants to enclose
a rectangular area and then divide it into four pens with fencing parallel to one side of
the rectangle. What is the largest possible total area of the four pens?
(a) Draw several diagrams illustrating the situation, some with shallow, wide pens and
some with deep, narrow pens. Find the total areas of these configurations. Does it
appear that there is a maximum area? If so, estimate it.
(b) Draw a diagram illustrating the general situation. Introduce notation and label the
diagram with your symbols.
(c) Write an expression for the total area.
(d) Use the given information to write an equation that relates the variables.
(e) Use part (d) to write the total area as a function of one variable.
(f) Finish solving the problem and compare the answer with your estimate in part (a).
(a) Here are three diagrams. The area of the one on the left is 12500 ft2 , the area of the
middle one is 12500 ft2 , and the area of the one on the left is 9000 ft2 . It seems that there
is a maximum and I would put it at around 13000 ft2 .
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(b) Let x be one side of the overall pen and y the other. The general situation is
(c) The total area is A = xy.
(d) The variables are related by 5x + 2y = 750.
(e) Since 5x + 2y = 750, we have y = 375 − 52 x. Thus we can write
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A( x ) = xy = 375x − x2 .
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(f) We find that A0 ( x ) = 375 − 5x. Setting A0 ( x ) = 0, we solve to find x = 75. Note that
for x < 75, A0 ( x ) > 0 and for x > 75, A0 ( x ) < 0. Thus x = 75 is a maximum of A by the
first derivative test. Furthermore, A(75) = 14062.5 ft2 , which is larger than my estimate
in part (a).
4.5.11. (a) Show that of all rectangles with a given area, the one with smallest perimeter
is a square.
(b) Show that of all rectangles with a given perimeter, the one with greatest area is a
square.
(a) Let x and y be the sides of the rectangle. We have A = xy, where A is constant.
We want to minimize P = 2x + 2y. Since xy = A, we have y = A/x. √
Thus we have
0
2
P( x ) = 2x + 2A/x. Setting P ( x ) = 0, we have 0√= 2 − 2A/x . Thus x = ± √A. Since side
lengths of rectangles
are positive, we have x = A. Note that for 0 < x < A, P0 ( x ) < 0
√
0
and for
√ x > A, P ( x ) > 0. The argument for this is similar to that in
√ problem
√ 4.5.3. Thus
x = A is a minimum of P. Furthermore, we have y = A/x = A/ A = A = x. Since
x = y in this case, the rectangle with smallest perimeter and area A is a square.
(b) Let x and y be the sides of the rectangle. We have P = 2x + 2y where P is constant.
Hence we have y = 12 P − x. We want to maximize A = xy. Plugging in for y we have
A( x ) =
P
x − x2 .
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Thus A0 ( x ) = P/2 − 2x. Setting P0 ( x ) = 0 and solving for x we get x = P/4. Note that
for 0 < x < P/4, A0 ( x ) > 0 and for x > P/4, A0 ( x ) < 0. Thus, by the first derivative
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test, x = P/4 is a maximum. In this case we have y = P/2 − x = P/2 − P/4 = P/4 = x.
Since x = y we have that of all rectangles with a given perimeter, the one with greatest
area is a square.
4.5.13. Find the points on the ellipse 4x2 + y2 = 4 that are farthest away from the point
(1, 0).
p
We want to maximize d = ( x − 1)2 + y2 given that 4x2 + y2 = 4. Notice that to have
4x2 + y2 = 4, we must have | x | ≤ 1. While we could solve for x or y and plug it in, it is
better not to in this case because we do not want to deal with taking plus or minus square
roots. Rather, notice that
q
q
q
d = ( x − 1)2 + y2 = x2 − 2x + 1 + y2 = 1 − 2x − 3x2 + 4x2 + y2
p
= 1 − 2x − 3x2 + 4
p
= 5 − 2x − 3x2 .
√
Thus we want to maximize d( x ) = 5 − 2x − 3x2 . Setting the derivative equal to 0 yields
−2 − 6x
0 = d0 ( x ) = √
.
2 5 − 2x − 3x2
Solving yields x = −1/3. For −1 < x < −1/3, −2 − 6x > 0, so d0 ( x ) > 0. Similarly,
for −1/3 < x < 1, we have d0 ( x ) < 0. Thus x = −1/3 is a maximum
of d by the first
√
2 + y2 = 4 for y, we find that y = ±2 1 − x2 , so at x = −1/3,
derivative
test.
Solving
4x
√
2
2
y = ±4 2/3. Thus the two √
points on the ellipse 4x
√ + y = 4 that are farthest away from
the point (1, 0) are (−1/3, 4 2/3) and (−1/3, −4 2/3).
4.5.18. Find the area of the largest rectangle that can be inscribed in the ellipse x2 /a2 +
y2 /b2 = 1.
In this case the area of the rectangle is A = 4xy, where ( x, y) is the coordinate of the corner
2
2
2 /b2 = 1 for y, we get
in the first
√ quadrant. We may assume a, b > 0. Solving x /a + y √
y = ±b 1 − x2 /a√2 . Since y lies in the first quadrant we have y = b 1 − x2 /a2 . Thus we
have A( x ) = 4bx 1 − x2 /a2 . Setting A0 ( x ) = 0 we find
0 = 4b
p
1 − x2 /a2 −
√
Multiplying by a2 1 − x2 /a2 gives
√
4bx2
a2 1 − x2 /a2
.
0 = 4ba2 (1 − x2 /a2 ) − 4bx2 = 4ba2 − 8bx2 ,
√
so x = a/ 2, where again we have chosen the positive value of the square root. Analysis
like those in previous problems
shows that A is maximized by this value of A. Thus the
√
maximum area is A( a/ 2) = 2ba.
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4.5.23. A cone shaped drinking cup is made by cutting out a sector of a paper circle of
radius R and joining the two edges. Find the maximum capacity of such a cup.
Let r be the radius of the circle at the top of the√cone shaped cup. The Pythagorean
theorem tells us that the height of the cone is h = R2 − r2 . Thus the volume is
V (r ) =
π p
π 2
r h = r 2 R2 − r 2 .
3
3
Setting the derivative equal to 0 we get
√
πr3
2πr R2 − r2
√
−
0 = V (r ) =
.
3
3 R2 − r 2
0
Notice that r = 0 is a solution, but in this case V (r ) = 0, which cannot be the maximum√(also, 0 is not in the domain of V 0 ). Solving with r 6= 0 yields the positive solution
r = 6R/3. Analysis similar to previous problems shows this is a maximum. Thus the
maximum capacity is
√
2πR3
V ( 6R/3) = √ .
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4.5.33. Find the equation of the line through (3, 5) that cuts off the least area from the
first quadrant.
Using the point slope form, the general equation for the line is y = m( x − 3) + 5. If m ≥ 0,
this line cuts off no area from the first quadrant, but this is a triviality, so assume m < 0.
Then we x intercept is at x = 3 − 5/m and the y intercept is at y = 5 − 3m. Notice that the
line cuts of a triangle with base the distance from the origin to the x intercept and height
the distance to the y intercept. Thus the area cut off is
A(m) = xy/2 = (5 − 3m)(3 − 5/m)/2 =
1
(30 − 25/m − 9m).
2
Setting A0 = 0, we get 0 = 25/m2 − 9. Solving for m we get m = −5/3, where we take
the negative value of the square root since we assumed m < 0. The first derivative test
shows this m minimizes A, so the equation of the line through (3, 5) that cuts off the least
area from the first quadrant is y = − 35 x + 10.
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4.5.37. A baseball team plays in a stadium that holds 55000 spectators. With tickets
priced at $10, the average attendance had been 27000. When ticket prices were lowered
to $8 the attendance rose to 33000.
(a) Find the demand function, assuming that it is linear.
(b) How should ticket prices be set to maximize revenue?
(a) Let p( x ) be the demand function. Then 10 = p(27000) and 8 = p(33000). Since p is
assumed to be linear, this gives a slope of
1
10 − 8
=−
.
27000 − 33000
3000
Thus using the point-slope formula we have
p( x ) = −
1
1
( x − 27000) + 10 = −
x + 19.
3000
3000
(b) The revenue is R( x ) = xp( x ) = 19x − x2 /3000. Setting R0 ( x ) = 0 gives 0 = 19 −
x/1500. Solving for x gives x = 28500. The same analysis as before shows that x is a
maximum. At this value the price is p(28500) = 9.5. Thus ticket prices should be set at
$9.5.
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