MATH 142–511, 516, 517
Spring 2010
Solutions for the sample problems for the Final
1. How many CDs would a record company have to make and sell to break even if the fixed
costs are $18000, variable costs are $5.20 per CD, and the CDs are sold to retailers for
$7.60 each?
SOLUTION. The cost function is C(x) = 18000 + 5.2x, the revenue function is R(x) =
7.6x. The company breaks even if R(x) = C(x)
18000 + 5.2x = 7.6x
18000
x=
= 7500
7.6 − 5.2
2. Trunsville Utilities uses the following rates to compute the monthly cost of natural gas
for residential customers. Write a piecewise definition for the cost of consuming x CCF
(cubic hundred feet) of natural gas.
Charges per month
$0.7675 per CCF for the first 50 CCF
$0.6400 per CCF for the next 150 CCF
$0.6130 per CCF for all over 200 CCF
SOLUTION. If 0 ≤ x ≤ 50, then C(x) = 0.7675x. Note, that C(50) = (50)(0.7675) =
38.375.
If 50 ≤ x ≤ 200, then C(x) = 38.375 + 0.64(x − 50) = 6.375 + 0.64x. C(200) =
6.375 + (0.64)(200) = 134.375
If x > 200, then C(x) = 134.375 + 0.613(x − 200) = 11.775 + 0.613x.
Thus

 0.7675x,
C(x) =
6.375 + 0.64x,

11.775 + 0.613x,
if 0 ≤ x ≤ 50
if 50 ≤ x ≤ 200
if x > 200
3. Find the domain of the function
SOLUTION.
 √
 x − 5,
1
,
f (x) =
 2x+5
5x,
if 5 ≤ x ≤ 10
if 10 < x ≤ 20
if x > 20
x−5 ≥0
x≥5
and 2x + 5 6= 0 or x 6= −2.5. Since −2.5 < 10, the domain of f is [5, ∞).
4. The revenue and cost functions for a particular product are given below. The cost and
revenue are given in millions of dollars, and x represents the number of units (in thousands).
R(x) = −64x2 + 4400x
C(x) = 250x + 5200
At what production level(s), rounded to the nearest whole unit, will the company break
even on this product?
SOLUTION. The company breaks even if R(x) = C(x).
−64x2 + 4400x = 250x + 5200
x1 ≈ 1, x2 ≈ 64
5. Solve the following equations for x:
(a) 26x = 8x
2 +1
2
26x = 8x +1
2
26x = (23 )x +1
2
26x = 23x +3
6x = 3x2 + 3
2x = x2 + 1
2
x − 2x + 1 = 0
(x − 1)2 = 0
x=1
(b) (5 − x)5 = (2x − 1)5
(5 − x)5 = (2x − 1)5
5 − x = 2x − 1
3x = 6
x=2
(c) 5x = 14
5x = 14
ln(5x ) = ln(14)
x ln 5 = ln 14
ln 14
x=
≈ 1.64
ln 5
(d) log2 (x + 2) + log2 (2) = log2 21
log2 (x + 2) + log2 (2) = log2 21
log2 2(x + 2) = log2 21
2(x + 2) = 21
21 − 4
x=
= 8.5
2
(e) log4 x = 3
log4 x = 3
x = 43 = 64
6. Suppose that $2500 is invested at 7% compounded quarterly. How much money will be
in the account in 5 years?
SOLUTION. A=$3536.95 (HINT: use TVM Solver for N = 4 ×5, I=7, PV=2500, P/Y=4
and solve for FV).
7. In its first 10 years the Gabelli Growth Fund produced an average annual return of 21.36%.
Assume that money invested in this fund continues to earn 21.36% compounded annually.
How long will it take money invested in this fund to double?
SOLUTION. t = 3.58 ≈ 4(years) (HINT: use TVM Solver for I=21.36, PV=1, FV=-2,
P/Y=1 and solve for N).
8. Evaluate each limit:
x2 − 3x + 2
2−1
1
(x − 2)(x − 1)
x−1
= lim
= lim
=
=
2
x→2 x + x − 6
x→2 (x − 2)(x + 3)
x→2 x + 3
2+3
5
(a) lim
2x3 + x2 + 5
2x3
2
=
lim
= lim
= −2
3
3
x→∞
x→∞
x→∞
x−x
−x
−1
(b) lim
(c) lim−
x→1
Since
|x − 1|
x−1
|x − 1| =
then
lim−
x→1
x − 1,
−(x − 1),
if x ≥ 1
if x < 1
|x − 1|
−(x − 1)
= lim−
= −1.
x→1
x−1
x−1
9. Determine where the function
f (x) =
1 + x,
6 − x,
if x ≤ 2
if x > 2
is continuous.
SOLUTION. If x < 2, then f (x) = 1 + x and f is continuous on its domain (−∞, 2).
If x > 2 then f (x) = 6 − x and f is continuous on its domain (2, ∞).
lim f (x) = lim+ (6 − x) = 6 − 2 = 4
x→2+
x→2
lim− f (x) = lim− (1 + x) = 1 + 2 = 3
x→2
x→2
Since lim+ f (x) 6= lim− f (x), then lim f (x) does not exist. Thus f (x) is discontinuous at
x→2
x = 2.
x→2
x→2
f (x) is continuous on (−∞, 2) ∪ (2, ∞).
10. Find all vertical and horizontal asymptotes for the function
f (x) =
(2x − 3)(x + 2)(x2 − 1)
(x − 4)(x + 3)x(x − 1)
SOLUTION.
(2x − 3)(x + 2)(x2 − 1)
(2x − 3)(x + 2)(x − 1)(x + 1)
(2x − 3)(x + 2)(x + 1)
f (x) =
=
=
(x − 4)(x + 3)x(x − 1)
(x − 4)(x + 3)x(x − 1)
(x − 4)(x + 3)x
Lines x = −3, x = 0, and x = 4 are the vertical asymptotes for f .
(2x − 3)(x + 2)(x + 1)
= 2, the equation of the horizontal asympSince lim f (x) = lim
x→∞
x→∞
(x − 4)(x + 3)x
tote is y = 2.
11. Let
p = 25 − 0.01x
and
C(x) = 2x + 9000
where 0 ≤ x ≤ 2500, be the price-demand equation and cost function, respectively, for
the manufacture of umbrellas.
(a) Find the exact cost of producing the 31st umbrella. Use the marginal cost to approximate the cost of producing the 31st umbrella.
SOLUTION. The exact cost of producing the 31st umbrella is C(31) − C(30) =
2(31) + 9000 − (2(30) + 9000) = $2.
The marginal cost function C ′ (x) = 2. C ′ (30) ≈ C(31) − C(30). C ′ (30) = 2.
(b) Find the marginal revenue and the marginal average revenue functions.
SOLUTION. The revenue R(x) = xp(x) = 25x − 0.01x2
The marginal revenue function is R′ (x) = 25 − 0.02x.
R(x)
25x − 0.02x2
The average revenue is R̄ =
=
= 25 − 0.01x.
x
x
The marginal average revenue is function R̄′ (x) = −0.01.
(c) Find the average profit per umbrella if 50 umbrellas is produced. Find the marginal
average profit at a production level of 50 umbrellas. Estimate the average profit per
umbrella if 51 umbrella is produced.
SOLUTION. The profit is P (x) = R(x) − C(x) = 25x − 0.01x2 − (2x + 9000) =
−0.01x2 + 23x − 9000.
P (x)
9000
The average profit is P̄ (x) =
= −0.01x + 23 −
.
x
x
9000
The marginal average profit is P̄ ′(x) = −0.01 + 2 .
x
The average profit per umbrella if 50 umbrellas is produced is P̄ (50) = −0.01(50) +
9000
= −157.5
23 −
50
The marginal average profit at a production level of 50 umbrellas is P̄ ′(50) = −0.01+
9000
= 3.59
(50)2
The average profit per umbrella if 51 umbrella is produced is P̄ (50) + P̄ ′ (50) =
−157.5 + 3.59 = −153.91
12. A bank offers a 10-year certificate of deposit (CD) that earns 4.15% compounded continuously.
(a) If $10000 is invested in this CD, how much will it be worth in 10 years?
SOLUTION. A = P ert , where P = principal, r = annual nominal interest rate
compounded continuously, t = time in years, A = amount at time t.
P = 10000, r = 0.0415, t = 10, A =?.
A = 10000e0.0415×10 = $15143.71
(b) How long will it take for the account to be worth $18000?
A = 18000, P = 10000, r = 0.0415, t =?
A = P ert
A
= ert
P
A
ln = ln ert
P
A
ln = rt
P
1
18000
1 A
ln
≈ 14
t = ln =
r P
0.0415 10000
13. A note will pay $25000 at maturity 10 years from now. How much should you willing to
pay for the note now if money is worth 5% compounded continuously?
SOLUTION. A = 25000, t = 10, r = 0.05, P =?.
P = Ae−rt
A = P ert
= 25000e−0.05×10 ≈ $15163.27
14. At what nominal rate compounded continuously must money be invested to double in 8
years?
SOLUTION. A = 2P , t = 8, r =?.
2P = P e8r
2 = e8r
ln 2 = ln (e8r )
ln 2 = 8r
ln 2
r=
≈ 8.66%
8
15. Find the equation of the tangent line to the graph of the function f (x) = ln(1 − x2 + 2x4 )
at the point where x = 1.
SOLUTION. The equation of the tangent line to the graph of the function f (x) = ln(1 −
x2 + 2x4 ) at the point where x = 1 is
y − f (1) = f ′ (1)(x − 1)
f (1) = ln(1 − 12 + 2(14 )) = ln 2.
(1 − x2 + 2x4 )′
−2x + 8x3
−2 + 8
6
=
, f ′ (1) =
= =3
2
4
2
4
1 − x + 2x
1 − x + 2x
1−1+2
2
Thus, the equation of the tangent line is y − ln 2 = 3(x − 1) or y = 3x − 3 + ln 2.
f ′ (x) =
16. Find the value(s) of x where the tangent line to the graph of the function y = 5ex
is horizontal.
2 −4x+1
SOLUTION. The tangent line is horizontal means that its slope is zero.
f ′ (x) = 5ex
2 −4x+1
(x2 − 4x + 1)′ = (2x − 4)5ex
2x − 4 = 0
x=2
2 −4x+1
=0
17. Find each derivative
(a)
(b)
(c)
(d)
√
d
d
d 1
1 d
1/4
4
log3 ( 4x3 + 5x + 7) = log3 (4x3 + 5x + 7 ) =
log3 (4x3 +5x+7) =
log (4x3 +
dx
dx
dx 4
4 dx 3
1
12x2 + 5
1 (4x3 + 5x + 7)′
=
5x + 7) =
4 (4x3 + 5x + 7) ln 3 4 (4x3 + 5x + 7) ln 3
d 1−2x3
3
3
8
=81−2x (1 − 2x3 )′ ln 8 =−6x2 81−2x ln 8.
dx
6x(x2 + 5)3 − 3(x2 + 5)2 (x2 + 5)′ (3x2 ) 6x(x2 + 5)3 − 3(x2 + 5)2 (2x)(3x2 )
3x2
d
=
=
=
dx (x2 + 5)3
(x2 + 5)6
(x2 + 5)6
6x(x2 + 5)2 [x2 + 5 − 3x2 ]
6x(5 − 2x2 )
=
(x2 + 5)6
(x2 + 5)4
d
[(x2 + x − 3)e2x+3 ] = (2x + 1)e2x+3 + e2x+3 (2x + 3)′ (x2 + x − 3) = (2x + 1)e2x+3 +
dx
e2x+3 (2)(x2 + x − 3) = (2x2 + 4x − 5)e2x+3
18. Given the price-demand equation
0.02x + p = 60
(a) Find the elasticity of demand E(p).
SOLUTION. First, let’s solve the equation 0.02x + p = 60 for x.
x=
60 − p
= 3000 − 50p.
0.02
E(p) =
−pf ′ (p)
,
f (p)
where f (p) = 3000 − 50p.
−pf ′ (p)
−p(−50)
p
E(p) =
=
=
f (p)
3000 − 50p
60 − p
(b) For which values of p is demand elastic?
SOLUTION. The demand is elastic if E(p) > 1.
p
>1
60 − p
p
−1>0
60 − p
60 − p
p
−
>0
60 − p 60 − p
p − 60 + p
>0
60 − p
2p − 60
>0
60 − p
p − 30
>0
60 − p
30 < p < 60
(c) If p = $10 and the price is increased by 5%, what is the approximate change in
demand?
10
= 0.2.
SOLUTION. E(10) =
60 − 10
(change is demand)=E(10)×(change in price)=(0.2)(5%)=1%
(d) If p = $40 and the price is decreased, will revenue increase or decrease?
40
SOLUTION. E(40) =
= 2 > 1, the demand is elastic. If the price is
60 − 40
decreased then the revenue will increase.
19. Find f ′′ (x) for the functions
(a) f (x) = x2 (2x3 − 5)4
SOLUTION. f ′ (x) = 2x(2x3 − 5)4 + (x2 )(4)(2x3 − 5)3 (2x3 − 5)′ = 2x(2x3 − 5)4 +
(x2 )(4)(2x3 − 5)3 (6x2 ) = 2x(2x3 − 5)4 + 24x4 (2x3 − 5)3
f ′′ (x) = 2(2x3 −5)4 +(2x)(4)(2x3 −5)3 (2x3 −5)′ +(24)(4x3 )(2x3 −5)3 +24x4 (3)(2x3 −
5)2 (2x3 −5)′ = 2(2x3 −5)4 +(2x)(4)(2x3 −5)3 (6x2 )+(24)(4x3)(2x3 −5)3 +24x4 (3)(2x3 −
5)2 (6x2 ) = 2(2x3 − 5)4 + 48x3 (2x3 − 5)3 + 96x3 (2x3 − 5)3 + 432x6 (2x3 − 5)2 =
2(2x3 − 5)4 + 144x3 (2x3 − 5)3 + 432x6 (2x3 − 5)2
6
2
− 3 = 2x−1 − 6x−3 .
x x
SOLUTION. f ′ (x) = 2(−1)x−2 − 6(−3)x−4 = −2x−2 + 18x−4
(b) f (x) =
f ′′ (x) = −2(−2)x−3 + 18(−4)x−5 = 4x−3 − 72x−5 .
20. Given the graph of the derivative f ′ (x) of the function y = f (x).
(a) Find the intervals on which f is increasing, decreasing.
SOLUTION. f is increasing on the interval where f ′ (x) > 0 and decreasing on the
interval where f ′ (x) < 0.
f is increasing on (a, e) ∪ (g, ∞).
f is decreasing on (e, g).
(b) Find x-coordinates of the critical points for the function f . x = e, and x = g.
(c) Find the intervals on which f is concave upward, concave downward.
SOLUTION. f is CU on the interval where f ′ is increasing and CD on the interval
where f ′ is decreasing.
f is CU on (b, c) ∪ (f, ∞)
f is CD on (a, b) ∪ (c, f ).
(d) Find x-coordinates of the inflection points for f . x = b, x = c, and x = f .
1
1
21. Given the function f (x) = x3 + x2 − 3x + 4.
3
2
(a) Find critical values of f (x).
f ′ (x) = x2 + x − 3 = 0
x1 = −2.3, x2 = 1.3
(b) Find intervals on which f (x) is increasing and decreasing.
To find intervals we have to constract the sigh chart for f ′ (x).
The sign chart for f ′ (x)
f is increasing on (−∞, −2.3) ∪ (1.3, ∞) and decreasing on (−2.3, 1.3).
(c) Find local extrema for f (x).
It follows from the sign chart that f has the local minimum at x = 1.3 and the local
maximum at x = −2.3.
(d) Find intervals on which f (x) is concave upward and concave downward.
f ′′ (x) = 2x − 1 = 0
x = −0.5
To find intervals we have to construct the sign chart for f ′′ (x).
The sign chart for f ′′ (x)
f is CU on (−0.5, ∞) and CD on (∞, −0.5).
(e) Find all inflection points of f (x).
f has the inflection point at x = −0.5
22. Find the absolute maximum and absolute minimum for the function f (x) =
the interval [−1, 2].
√
9 − x2 on
SOLUTION.
1
1
f ′ (x) = (9 − x2 )1/2−1 (9 − x2 )′ = (9 − x2 )−1/2 (−2x) = −x(9 − x2 )−1/2 = 0
2
2
x=0
√
√
9
−
1
=
f (−1) =
√ 8 ≈ 2.83
√
f (0) = √9 − 0 = √9 = 3 absolute maximum value
f (2) = 9 − 4 = 5 ≈ 2.24 absolute minimum value
23. Find the absolute maximum and minimum for the function f (x) =
SOLUTION.
f ′ (x) =
x2 − 1
.
x2 + 1
2x(x2 + 1) − 2x(x2 − 1)
4x
= 2
=0
2
2
(x + 1)
(x + 1)2
x=0
To classify the critical value x = 0 we have to use the Second Derivative Test.
f ′′ (x) =
4(x2 + 1)2 − (4x)(2)(x2 + 1)(x2 + 1)′
4(x2 + 1)2 − 8x(x2 + 1)(2x)
=
(x2 + 1)4
(x2 + 1)4
′′
f (0) = 4 > 0
Function f has the absolute minimum at x = 0.
24. What are the dimensions of the rectangular field of 20000 square feet that will minimize
the cost of fencing if one side costs three times as much per unit length as the other three?
SOLUTION. Let x be the length of the rectangle and y be the width of the rectangle.
The area of the rectangle is A = xy = 20000. Then
20000
.
y=
x
The fence along three sides is to be made of material
that costs $p per foot. The material of the fourth side
costs $3p per foot.
Then the cost of the fence is
C(x) = (2y + x)p + 3px = 4px + 2py = 2p(2x + y) = 2p(2x +
20000
)=0
C ′ (x) = 2p(2 −
x2
10000
1−
=0
2
x
x2 = 10000
20000
x = 100, y =
= 200
100
20000
)
x
25. A 300-room hotel in Las Vegas is filled to capacity every night at $100 a room. For each
$2 increase in rent, 5 fewer rooms are rented. If each rented room costs $10 service per
day, how much should be management charge for each room to maximize gross profit?
What is the maximum gross profit?
SOLUTION. Let x be the number of $2 increase in rent. Then
the rent is 100 + 2x
the number of rooms rented is 300 − 5x
the profit is P (x) = (100+2x−10)(300−5x) = (90+2x)(300−5x). We have to maximize
profit.
P ′ (x) = 2(300 − 5x) − 5(90 + 2x) = 150 − 20x = 0
x = 7.5
To maximize the profit management should charge 100 + (2)(7.5) = $115 for each room.
The maximum profit is P (7.5) = (90 + (2)(7.5))(300 − (5)(7.5)) = $27652.50
26. Find the following indefinite integrals:
u = √x = x1/2
Z
√
1
√ e x dx = du = 1/2x−1/2 dx
(a)
x
x−1/2 dx = 2du
Z
Z
√
= eu (2du) = 2 eu du = 2eu +C = 2e x +C.
u = x4 − 12x + 3
x −3
(b)
dx = du = (4x3 − 12)dx = 4(x3 − 3)dx
4
x − 12x + 3
(x3 − 3)dx = 1/4du
1
1
ln |u| + C = ln |x4 − 12x + 3| + C
4
4
Z
(c)
Z
(d)
Z
3
u = ln x
(ln x)2
dx = 1
du = dx
x
x
Z
u3
(ln x)3
+C =
+C
= u2 du =
3
3
u = 5 − 2x2
x
du = −4xdx
dx
=
(5 − 2x2 )5
xdx = −1/4 du
1
(5 − 2x2 )−4 + C
16
Z
11
=
du =
4u
Z
−4
= − 1 u−5du = − 1 u + C =
4
4 −4
u=3−x Z
Z
Z
3−u
x
−1/2
√
√ du = − (3−u)u
dx = x = 3 − u = −
du = − (3u−1/2 −
(e)
u
3−x
dx = −du u1/2 u3/2
2
2
u1/2 )du = −3
+
+ C = −6u1/2 + u3/2 + C = −6(3 − x)1/2 + (3 − x)3/2 + C
1/2 3/2
3
3
Z
u = 2x − 1 Z
1Z
u
+
1
1
1
x=
2x−1
(f)
e
dx = eu du = eu + C = e2x−1 + C
=
2
2
2
2
dx = 1 du 2
27. The weekly marginal revenue from the sale of x pairs of tennis shoes is given by
R′ (x) = 40 − 0.002x +
200
,
x+1
R(0) = 0,
where R(x) is revenue in dollars. Find the revenue function R(x). Find the revenue from
the sale of 1000 pairs of shoes.
Z
Z 1
200
2
dx = 40x − 0.01x + 200
dx =
SOLUTION. R(x) =
40 − 0.02x +
x+1
x+1
Z
x+1 =u 1
2
2
dx = du = 40x − 0.01x + 200 u du = 40x − 0.01x + 200 ln |u| + C = 40x −
0.01x2 + 200 ln |x + 1| + C
R(0) = (40)(0) − (0.01)(0) + 200 ln |0 + 1| + C = C = 0
Thus, R(x) = 40x − 0.01x2 + 200 ln |x + 1|.
R(1000) = (40)(1000) − 0.01(1000)2 + 200 ln(1001) = 40381.75
28. Calculate the following Riemann sum for the function f (x) = x2 − 9x + 7. Partition the
interval [−1, 3] into four subintervals of equal length. For each subinterval [xk , xk+1 ] let
ck be the midpoint.
3 − (−1)
= 1, the partition points are
SOLUTION. The length of subintervals is ∆x =
4
x0 = −1, x1 = −1 + 1 = 0, x2 = 0 + 1 = 1, x3 = 1 + 1 = 2, and x4 = 3. The midpoint
of the interval [−1, 0] is c1 = (−1 + 0)/2 = −0.5, the midpoint of the interval [0, 1] is
c2 = (0 + 1)/2 = 0.5, the midpoint of the interval [1, 2] is c3 = (1 + 2)/2 = 1.5, and the
midpoint of the interval [2, 3] is c4 = (2 + 3)/2 = 2.5.
The Riemann sum S4 = f (c1 )∆x + f (c2 )∆x + f (c3 )∆x + f (c4 )∆x = f (−0.5)(1) +
f (0.5)(1) + f (1.5)(1) + f (2.5)(1) = 11.75 + 2.75 + (−4.25) + (−9.25) = 1
29. Evaluate each of the definite integrals:
(a)
RA
32(x2 + 1)7 x dx (A > 0)
0
RA
2
7
SOLUTION. 32(x + 1) x dx = 0
2((A2 + 1)8 − 18 ) = 2(A2 + 1)8 − 2
x2 + 1 = u
du = 2xdx
1
xdx = du
2
0 → 02 + 1 = 1
A → A2 + 1
AR2 +1
32u7 21 du = 16
=
1
A2 +1
u8 8 1
=
Z2 x2
dx, (0 < B < 2)
(b)
5x − 4 √
4
x3
B
Z2
Z2 Z2
x2
2−3/4
(5x − 4x5/4 )dx =
5x − 4 √
(5x − 4x
)dx =
SOLUTION.
dx =
4
3
x
B
B
B
2
2
x
5 2
x1+5/4
16
5
16
5 −4
= (2 − B 2 ) − (29/4 − B 9/4 ) = 1.5434 − B 2 + B 9/4
2
1 + 5/4 B 2
9
2
9
30. Calculate the definite integral
below.
Rd
b
f (x)dx by referring to the figure and list of area values
if of A = 1.4, area of B = 2.4, area of C = 3.1, area of D = 2.1.
Rd
SOLUTION. f (x)dx=-area(B)-area(C)+area(D)=-2.4-3.1+2.1=-3.4
b
31. Find the average value of the function f (x) = 4x − 3x2 over the interval [−2, 2].
Z2
1
SOLUTION. fave =
(4x − 3x2 )dx = −4
2 − (−2)
−2
32. Find the area of the region bounded by:
(a) y = 3 − x2 , y = 2x2 − 4x
SOLUTION.
y
y = 3 − x2
1.87
−0.54
x
y = 2x2 − 4x
Area =
1.87
R
−0.54
(3 − x2 − (2x2 − 4))dx = 10.17
(b) y = x3 , y = 4x
SOLUTION.
y
y = 4x
y = x3
−2
2
x
R2
Area = 2 (4x − x3 )dx = 8
0
(c) y = −x2 − 2x, y = 0, x = −2, x = 1.
SOLUTION.
y
y = −x2 − 2x
−2
Area=
Z0
−2
(−x2 − 2x)dx +
Z1
(x2 + 2x)dx =
1
x
8
3
0
33. Find the consumers’ surplus and producers’ surplus at the equilibrium price level for the
given price-demand and price-supply equations.
p = D(x) = 70 − 0.2x
p = S(x) = 13 + 0.0012x2
Round all values to the nearest integer.
SOLUTION. The equilibrium quantity is the solution to the equation
D(x) = S(x)
70 − 0.2x = 13 + 0.0012x2
950
x1 = −
, x2 = 150
3
Since x1 < 0, then x̄ = 150.
Then the consumers’ surplus is
CS =
Zx̄
p̄ = 70 − 0.2(150) = 40
Z150
(D(x) − p̄)dx = (70 − 0.2x − 40)dx = 2250
0
0
and the producers’ surplus is
PS =
Zx̄
0
Z150
(p̄ − S(x))dx = (40 − (13 + 0.0012x2 ))dx = 2700
0
10
. Find f (2, −3) − 4g(−1, 2).
+ 4y
10
203
SOLUTION. f (2, −3) − 4g(−1, 2) = 2(2) − 3(−3) + 14 − 4
=
2
(−1) + 4(2)
9
34. Let f (x, y) = 2x − 3y + 14 and g(x, y) =
x2
35. Weston Publishing publishes a deluxe edition and a standard edition of its English language dictionary. Weston’s management estimates that the number of deluxe editions
demanded is x copies/day and the number of standard editions demanded is y copies/day
when the unit prices are
p = 20 − 0.005x − 0.001y
q = 15 − 0.001x − 0.003y
dollars, respectively. Find the daily total revenue function R(x, y). Evaluate R(7, 3).
SOLUTION. R(x, y) = xp(x)+yq(y) = x(20−0.005x−0.001y)+y(15−0.001x−0.003y) =
20x − 0.005x2 − 0.002xy + 15y − 0.003y 2
R(7, 3) = 20(7) − 0.005(7)2 − 0.002(7)(3) + 15(3) − 0.003(3)2 = 184.686
36. Find the cross-section of the surface z = 10x + 4xy + 15y 2 − 6x2 + 5 produced by the
cutting it with the planes x = 4, y = 2.
SOLUTION. If we cut the surface by the plane x = 4 we get
z = 10(4) + 4(4)y + 15y 2 − 6(4)2 + 5 = −51 + 16y + 15y 2 (parabola)
The cross-section of the surface by the plane y = 2 is
z = 10x + 4x(2) + 15(2)2 − 6x2 + 5 = 65 + 18x − 6x2 (parabola)
37. Find Find fx , fy , fxx , fxy , and fyy for the functions:
p
(a) f (x, y) = 2x − y 2 = (2x − y 2)1/2
SOLUTION.
1
1
fx = (2x − y 2 )1/2−1 (2x − y 2)′x = (2x − y 2 )−1/2 (2) = (2x − y 2 )−1/2
2
2
1
1
fy = (2x − y 2)1/2−1 (2x − y 2 )′y = (2x − y 2 )−1/2 (−2y) = −y(2x − y 2)−1/2
2
2
1
1
fxx = − (2x − y 2 )−1/2−1 (2x − y 2 )′x = − (2x − y 2)−3/2 (2) = −(2x − y 2 )−3/2
2
2
1
1
fxy = − (2x − y 2 )−1/2−1 (2x − y 2)′y = − (2x − y 2)−3/2 (−2y) = y(2x − y 2)−3/2
2
2
1
′
2 −1/2
2 −1/2 ′
2 −1/2
fyy = −(y)y (2x − y )
− y((2x − y )
)y = −(2x − y )
−y −
(2x −
2
y 2 )−1/2−1 (2x − y 2 )′y = −(2x − y 2)−1/2 − y 2(2x − y 2 )−3/2
√
(b) f (x, y) = ex y
SOLUTION.
√
√
√
√ √
fx = ex y (x y)′x = yex y = y 1/2 ex y
√
1
√
(x y)′y = xy −1/2 ex y
2
√
√
√
= y 1/2 ex y (x y)′x = yex y
fy = ex
fxx
√
y
fxy = (y 1/2 )′y ex
1 x√y
xe
2
√
y
+ y 1/2 (ex
√
y ′
)y
=
√
√
1 −1/2 x√y
√
1
y
e
+ y 1/2 ex y (x y)′y = y −1/2 ex y +
2
2
√
√
√
√
1
1
1
1
√
−1/2 ′ x y 1
−1/2 x y ′
y −1/2−1 ex y + xy −1/2 ex y (x y)′y ==
)y e + xy
(e )y = x −
fyy = x(y
2
2
2
2
2
1 −3/2 x√y 1 2 −1 x√y
1 −3/2 x√y 1 −1/2 x√y 1 −1/2
e
e
xy
= − xy
e
+ xy
+ xy e
− xy
4
2
2
4
4
(c) f (x, y) = ln(x3 − y 2 )
SOLUTION.
(x3 − y 2 )′x
3x2
fx =
=
x3 − y 2
x3 − y 2
fy =
(x3 − y 2)′y
2y
=− 3
3
2
x −y
x − y2
fxx =
6x(x3 − y 2 ) − 3x2 (3x2 )
6x(x3 − y 2 ) − 9x4
(3x2 )′x (x3 − y 2 ) − (x3 − y 2 )′x 3x2
=
=
(x3 − y 2 )2
(x3 − y 2)2
(x3 − y 2 )2
fxy
(3x2 )′y (x3 − y 2 ) − (x3 − y 2 )′y 3x2
−(2y)(3x2 )
6x2 y
=
=
=
−
(x3 − y 2 )2
(x3 − y 2)2
(x3 − y 2 )2
fyy
(2y)′y (x3 − y 2) − (x3 − y 2)′y (2y)
2(x3 − y 2 ) − (−2y)(2y)
2(x3 − y 2) + 4y 2
=
−
=
−
=−
(x3 − y 2 )2
(x3 − y 2)2
(x3 − y 2 )2
38. Find the local extrema for the function f (x, y) = −x2 − y 2 + 2x + 4y + 5.
SOLUTION.
fx = −2x + 2 = 0
x=1
fy = −2x + 4 = 0
y=2
The critical point is (1, 2).
fxx = −2 = A < 0
fxy = 0 = B
fyy = −2 = C
AC − B 2 = (−2)(−2) − 02 = 4 > 0
Since A < 0 and AC − B 2 > 0, then f has the local maximum at (1, 2).
39. Average global temperatures from 1885 to 2005 are given in the table
Year
1885
1895
1905
1915
1925
1935
1945
o
F
56.65
56.64
56.52
56.57
56.74
57.00
57.13
Year
1955
1965
1975
1985
1995
2005
o
F
57.06
57.05
57.04
57.36
57.64
58.59
(a) Find the least squares line for the data, using x = 0 for 1885.
SOLUTION. The equation of the regression line is
y = 0.0121x + 56.379
(b) Use the least squares line to estimate the average global temperature in 2085.
SOLUTION. The average global temperature in 2085 is
y(200) = 0.0121(200) + 56.379 = 58.799