MATH 180: ELEMENTS OF CALCULUS I
Exam #3 – Solutions
W. Stalin Rios
Wednesday, June 16, 2014
1
Optimization
1. Find the absolute maximum value and the absolute minimum value of
1
f (x) = x +
x
on the interval [0.25, 2]
We first find the derivative f 0 (x) = 1 − 1/x2 and solve for the critical points.
0 = 1 − 1/x2 ⇒ x = ±1. Since x = −1 is outside the given domain, we only consider x = 1. We
now evaluate f (x) at the critical points and end points: f (0.25) = 4.25, f (1) = 2, and f (2) = 2.5.
Therefore, the absolute maximum is f (0.25) = 4.25 and the absolute minimum is f (1) = 2.
Maximizing Revenue
2. The quantity demanded each month of the Sicard wristwatch is related to the unit price
by the equation
10
(0 ≤ x ≤ 10)
p=
0.25x2 + 1
where p is measured in dollars and x is measured in units of a thousand. To yield a
maximum revenue, how many watches must be sold?
Hint: R(x) = px.
First, we define the revenue function: R(x) = px =
revenue function:
R0 (x) =
10x
. We now find the marginal
0.25x2 + 1
10(0.25x2 + 1) − 0.5x(10x)
10 − 2.5x2
=
(0.25x2 + 1)2
(0.25x2 + 1)2
Solving R0 (x) = 0 to find critical points yields x = ±2. Since x = −2 is outside the given
domain, we only consider x = 2.
We now evaluate R0 (x) at some test points on both sides of x = 2, thus R0 (0) = 10 ⊕ and
R (10) = (10 − 250)/⊕ = . Hence, R(x) has a maximum value at x = 2.
0
Therefore, in order to maximize the revenue of Sicard wristwatches, 2000 watches must be
sold.
1
Minimizing Costs
3. For its beef stew, Betty Moore Company uses aluminum containers that have the form of
right circular cylinders. Find the radius and height of a container if it has a capacity of
16π in3 and is constructed using the least amount of metal.
Hint: The volume of a right circular cylinder is given by the area of its base times the
height. V = πr2 h
The volume of such right cylinder is V = πr2 h = 16π ⇒ h = 16/r2 . Then, in order to use
the least amount of the material we have to minimize the surface area. Hence, SA = 2πr2 +2πrh.
Substituting h into SA yields f (x) = 2πr2 + 32π/r. The first derivative of the surface area
is f 0 (x) = 4πr − 32π/r2 and from f 0 (x) = 0 we find the critical point at r = 2.
Therefore, if the radius is 2in and the height is 4in, Betty Moore Company would then
construct right circular containers using the least amount of metal.
2
Exponential and Logarithmic Differentiation
4. Find the derivative of each function:
√ (a) f (x) = e2x ln 3x
First notice that f (x) = e2x ln(3x)1/2 =
1 2x
e ln(3x)
2
Now, using the product rule, we find
2
1 3
f 0 (x) = e2x ln(3x) +
e2x
2
2 3x
e2x
= e2x ln(3x) +
2x
(b) y = (x2 − 1)3 (2x + 6)2 (x − 5)
Hint: use logarithmic differentiation.
Taking the natural logarithm of both sides yields
ln(y) = ln (x2 − 1)3 (2x + 6)2 (x − 5)
= ln(x2 − 1)3 + ln(2x + 6)2 + ln(x − 5)
= 3 ln(x2 − 1) + 2 ln(2x + 6) + ln(x − 5)
Now, taking the derivative of both sides we find
y0
2x
2
1
=3
+2
+
y
x2 − 1
2x + 6
x−5
6x
4
1
= 2
+
+
x − 1 2x + 6 x − 5
2
Therefore,
6x
4
1
y0 = y
+
+
x2 − 1 2x + 6 x − 5
4
1
6x
= (x2 − 1)3 (2x + 6)2 (x − 5)
+
+
x2 − 1 2x + 6 x − 5
3
Integration
5. Find f (x) by solving the initial value problem
f 0 (x) = ex − 2x ;
f (0) = 2
Integrating both sides of the given equation yields
Z
Z
f 0 (x) dx = (ex − 2x) dx
f (x) = ex − x2 + C
Then, given than f (0) = 2 we solve for C as follow 2 = f (0) = e0 − 02 + C ⇒ C = 1.
Finally, we obtain f (x) = ex − x2 + 1
6. Solve the indefinite integral
Z 3ex − 3e−x
2
−
x(ln x)2
(ex + e−x )2/3
dx
First, we can separate the integrals and bring the constants outside
Z
Z
ex − e−x
1
dx
−
3
dx
2
2
x
x(ln x)
(e + e−x )2/3
To solve the first integral we can use the substitution u = ln(x) where du =
Z
2
1
dx = 2
x(ln x)2
Z
1
du = 2
u2
Z
u−2 du = 2
dx
so that
x
u−1
2
+C =− +C
−1
u
The second integral can be solved by substituting u = ex + e−x where du = (ex − e−x )dx
so that
Z
Z
Z
ex − e−x
1
u1/3
−2/3
3
+ C = 9u1/3 + C
dx
=
3
du
=
3
u
du
=
3
1/3
(ex + e−x )2/3
u2/3
Finally, substituting back in terms of x the solution to the indefinite integral is
Z 1/3
2
3ex − 3e−x
2
−
− 9 ex + e−x
+C
dx = −
2
x
−x
2/3
x(ln x)
ln(x)
(e + e )
3