MATH 211 FINAL EXAM SOLUTIONS 5/4/99
1. (30 points) Find the derivative of each of these functions. You do not need to simplify your answers.
(a) (p108 #29)f(x) = x
3+3
x=1
3x+ 3x−1f0(x) = 1
3−3x−2
(b) (p215 #15)f(x) = 2(x3−1)(3x2+ 1)4f0(x) = 2(3x2)(3x2+ 1)4+ 2(x3−1)(4)(3x2+ 1)3(6x)
(c) f(x) = ln(x4−ex)f0(x) = 4x3−ex
x4−ex
(d) f(x) = ex2−x
ex+ 1 f0(x) = (ex2−x)(2x−1)(ex+ 1) −(ex2−x)(ex)
(ex+ 1)2
2. (a) (7 pts; p96 #23) lim
x→6
x2−6x
x2−5x−6= lim
x→6
(x−6)(x)
(x−6)(x+ 1) = lim
x→6
x
x+ 1 =6
7
(b) (8 pts) lim
x→5
x2−10x+ 25
x2−25 = lim
x→5
(x−5)(x−5)
(x−5)(x+ 5) = lim
x→5
x−5
x+ 5 =0
10 = 0
3. (20 pts; p156, Ex 2) Using the techniques of calculus, sketch the graph of the function f(x) = x3−3x2+ 5.
On the graph, indicate all relative extreme points (relative maximum and relative minimum points) and all
points of inflection.
f0(x)=3x2−6x f00(x) = 6x−6 Setting f0(x) = 0 gives the potential extreme points x= 0, x= 2.
Since f00(0) = −6 and f00(2) = 6, there is a local maximum at (0,5) and a local minimum at (2,1). Setting
f00(x) = 0 shows that (1,3) is a point of inflection. You should also plot (−1,1) and (3,5) on the graph.
4. (5 pts) Let P(t) = 500 −100e−5t. When t= 2, is P(t) increasing or decreasing? Explain your answer.
We have P0(t) = −100e−5t(−5), so P0(2) = 500
e10 >0, and so P(t) is increasing when t= 2.
5. (10 pts) Find an equation for the line tangent to the curve y= 2x(x−4)6at x= 5.
When x= 5, y= 2·5(1)6= 10. y0= 2(x−4)6+(2x)(6)(x−4)5When x= 5, y0= 2(1)6+2·5·6(1)5= 62.
The equation of the tangent line is y= 62(x−5) + 10.
6. (15 pts; p201 #56) A closed rectangular box with a square base is to be constructed using two different
types of wood. The top is made of wood costing $3 per square foot, and the remainder is made of wood
costing $1 per square foot. Suppose that $48 is available to spend. Find the dimensions of the box of the
greatest volume that can be constructed.
Solution: If the box has a base that is xfeet by xfeet, with height h, then the problem is to maximize the
volume V=x2h(this is the objective equation). The cost of the box is 3x2+x2+ 4xh = 48, representing
the cost of the top, the bottom, and the four sides, respectively. This gives the constraint equation, which
we can solve for hin terms of x.
Objective: V=x2hConstraint: 3x2+x2+ 4xh = 48 x2+xh = 12 h=12 −x2
x
V=x2h=x212 −x2
x=x(12 −x2) = 12x−x3.V0(x) = 12 −3x2V00(x) = −6x
Setting V0(x) = 0 we get 12 −3x2= 0, or x2= 4, so x=±2.
We have V00(2) = −12, so V(x) is concave down at x= 2, showing that x= 2 does lead to a maximum value
for the area. Substituting x= 2 in the constraint equation h=12 −x2
xgives h= 4.
7. (12 pts) Ten grams of a certain radioactive material decays to three grams in five years. What is the
half-life of the radioactive material?
We use the equation P(t) = P0e−λt, with P0= 10, the initial amount.
P(5) = 3 10e−λ·5= 3 e−λ·5=.3−5λ= ln(.3) λ=−1
5ln(.3) = −.2 ln(.3)
