Review Questions 3
econ 11a
ams 11a
Solutions
The linear approximation formulas: If y = f (x) is differentiable at x = x0 , then:
f (x) ≈ f (x0 ) + f 0 (x0 )(x − x0 )
OR
∆f ≈ f 0 (x0 ) · ∆x.
These approximations are accurate when x − x0 = ∆x is sufficiently close to 0.†
1. Compute the derivatives, using the rules of differentiation, (no limits necessary).
a. f (x) = 3x5 − 2x4 + 5x2 − 3x − 2,
b. y =
√
3
x2 −
t2 + 3t − 1
,
3t4 + 5
c. g(t) =
d. u =
√
2
= x2/3 − 2x−3 ,
x3
g 0 (t) =
f 0 (x) = 15x4 − 8x3 + 10x − 3.
dy
2
= x−1/3 + 6x−4 .
dx
3
(2t + 3)(3t4 + 5) − (t2 + 3t − 1)(12t3 )
.
(3t4 + 5)2
v(v 2 + 3v + 5) = v 5/2 + 3v 3/2 + 5v 1/2 ,
du
5
9
5
= v 3/2 + v 1/2 + v −1/2 .
dv
2
2
2
e. y = (x2 + x + 1)3 = x6 + 3x5 + 6x4 + 7x3 + 6x2 + 3x + 11,
y 0 = 6x5 + 15x4 + 24x3 + 21x2 + 18x + 3.
Or, use the chain rule:
y 0 = 3(x2 + x + 1)2 · (2x + 1). (As an algebra exercise, verify that the two versions
of y 0 are in fact equal.)
In the last two problems, the chain rule is your only option...
f. h(x) =
√
3
2x2 + 3x − 5 = (2x2 + 3x − 5)1/3 ,
1
h0 (x) = (2x2 + 3x − 5)−2/3 · (4x + 3).
3
g. l(s) = (s2 + 2)3/2 · (4s + 7),
dl
= 32 (s2 + 2)1/2 · 2s · (4s + 7) + 4(s2 + 2)3/2 = (s2 + 2)1/2 · (16s2 + 21s + 8).
ds
†
What constitutes sufficiently close to 0 depends on the function, the point x0 , and how accurate
we would like the approximation to be. In practice, you may assume that x − x0 is always close
enough to 0 when you are asked to use the approximation formula.
1
2. Compute the derivative of y = (x2 + 1)(x2 − 2x + 2) in two different ways.
Product Rule: y 0 = 2x(x2 − 2x + 2) + (x2 + 1)(2x − 2) = 4x3 − 6x2 + 6x − 2.
Expand, then differentiate: y = x4 −2x3 +3x2 −2x+2 =⇒ y 0 = 4x3 −6x2 +6x−2.
3. Find the equation of the tangent line to the graph y =
√
1
x − √ at the point
x
(1, 0).
The slope of the tangent line is given by y 0 (1), so first we compute
1
1 −1/2
1 −1/2
0
− −
x−3/2 =
x
+ x−3/2 .
y = x
2
2
2
1 −1/2
Then we evaluate y 0 (1) =
1
+ 1−3/2 = 1. And finally, we use the point slope
2
formula:
y − 0 = 1 · (x − 1) =⇒ y = x − 1
4. The demand equation for a firm’s product is p = 200 − 0.3q 2/3 , where p is the
price, measured in $100s, and q is output (=demand) measured in 1000s of units.
dr
= 200 − 0.5q 2/3 and
a. r = pq = 200 − 0.3q 2/3 · q = 200q − 0.3q 5/3 , so
dq
dr 0
r (1000) =
= 200 − (0.5)10002/3 = 200 − 50 = 150.
dq q=1000
b. Since output is measured in 1000s of units, it follows that if output is 1,000,000
units, then q = 1000, and if output is 1,000,150 units, then q = 1000.15. Thus
the change in output in this case is ∆q = 0.15. Now, using the approximation
formula gives
!
dr · ∆q = 150 · (0.15) = 22.5,
∆r ≈
dq q=1000
so when output increases from 1,000,000 to 1,000,150, the firm’s revenue increases by about $2,250, (since revenue, like price, is measured in $100s).
5. A firm’s demand function is p = 300 − 0.4q, and their cost function is c =
0.05q 2 + 30q + 1000. Find the level of output, q, for which marginal revenue equals
marginal cost.
2
The revenue function is r = pq = 300q − 0.4q 2 , so
dr
= 300 − 0.8q. Marginal cost is
dq
dc
= 0.1q + 30, so marginal revenue equals marginal cost when
dq
300 − 0.8q = 0.1q + 30 =⇒ 270 = 0.9q =⇒ q = 300
6. The national savings function for Slugsylvania is
S=
2Y 2 + 5Y + 7
,
11Y + 111
where Y is annual income and S is annual savings, and both are measured in $ billions.
a.
(4Y + 5)(11Y + 111) − 11(2Y 2 + 5Y + 7)
dS
=
, so
dY
(11Y + 111)2
dS 45 · 221 − 11 · 257
dC dS =
≈ 0.146 and
= 1−
≈ 0.854.
dY Y =10
2212
dY Y =10
dY Y =10
b. To figure out the proportion of income that is saved when Y is very large, we
can use the same sort of reasoning that we do to compute limits at infinity.
Specifically, when Y is large, 2Y 2 is much larger than 5Y + 7, and 11Y is much
larger than 111, so for large values of Y
S=
2Y 2
2
2Y 2 + 5Y + 7
≈
= Y.
11Y + 111
11Y
11
From this we conclude that when Y is very large, approximately
dollar of income is saved.
2
11
of every
We can arrive at the same conclusion in another way, by investigating what
happens to the marginal propensity to save as Y grows very large. Too do this,
we first need to simplify the expression we found in part a. for dS/dY ,
(4Y + 5)(11Y + 111) − 11(2Y 2 + 5Y + 7)
22Y 2 + 444Y + 632
dS
=
=
.
dY
(11Y + 111)2
121Y 2 + 2442Y + 12321
It follows therefore that
dS
22Y 2 + 444Y + 632
22Y 2
2
=
= lim
=
lim
.
Y →∞ dY
Y →∞ 121Y 2 + 2442Y + 12321
Y →∞ 121Y 2
11
lim
This means that when Y is very large, approximately 2/11 of every additional
dollar of income is saved. This has the same implication as before, namely when
2
Y is very large then S(Y ) ≈ 11
Y , so we can draw the same conclusion as before.
3
7. The production function for ACME Widgets is given by
q = (5l + 4)2/3 ,
where q is the firm’s weekly output, measured in 1000s of widgets and l is the
firm’s weekly labor input, measured in $1000’s. The firm’s marginal revenue
function is given by
dr
= 80q −1/4 ,
dq
where revenue and marginal revenue are measured in $1000’s.
dq
10
dq 5
10 1
−1/3
a. Marginal product of labor:
= (5l + 4)
, so
· = .
=
dl
3
dl l=12
3 4
6
For marginal revenue product, we use the chain rule (see section 11.5):
dr dq
dr
=
· .
dl
dq dl
When l = 12, we have q(12) = 642/3 = 16, so
dr dr dq 5
100
=
·
= 80 · 16−1/4 · =
≈ 33.333.
dl l=12
dq q=16 dl l=12
6
3
dq 5
5
b. ∆q ≈
≈ 0.416667. In other words, output will
· ∆l = · 0.5 =
dl l=12
6
12
increase by about 417 widgets a week.
dr 100
50
c. ∆r ≈
· ∆l =
· 0.5 =
≈ 16.667. In other words, revenue will
dl l=12
3
3
increase by about $16,667 a week.
4