Winter 2005 Test 2 Solutions

Math 112, Calculus I
Test 2 Solutions
1. (20 points) Suppose you know that for a function f,
f(5) = 1 and f ′ (x) = 3√ x 2 + 2.
(a) Use a linear approximation to estimate f(5.4).
Solution: The linearization of f(x) has the form L(x) = f(a) + f ′ (a)[x − a]. We
wish to estimate f(5.4) and all we know is that f(5) = 1 and f ′ (x) =
3√ x 2 + 2, so
let’s use a = 5 in our approximation. Hence we find L(x) = f(5) + f ′ (5)[x − 5] =
[1] + [ 3√ (5) 2 + 2](x − 5) = 1 + [3](x − 5), so our estimation of f(5.4) is
f(5.4) ≈ L(5.4) = 1 + 3[(5.4) − 5] = 2.2
(b) Is your approximation of f(5.4) too large or too small? Justify your answer.
Solution: Our linearization above is the equation of the tangent line to the curve
y = f(x) at x = 5, which will lie below the curve itself if the curve is concave up at x = 5
and above if the curve is concave down at x = 5. To determine the concavity of y = f(x)
at x = 5, we compute the second derivative f ′′ (x) = d
dx [ 3√ x 2 + 2] = 1 3 (x2 + 2) −2/3 (2x)
and evaluate it at x = 5: f ′′ (5) = 1 3 [(5)2 +2] −2/3 [2(5)] = 10
27
. As this is positive, y = f(x)
is concave up at x = 5, so the tangent line lies beneath the curve itself and thus our
approximation in part (a) was too small.
2. (20 points) The formula for the volume of a right circular cone is V = 1 3 πr2 h.
(a) Write a formula that estimates the change that occurs in the volume of a right circular
cone when the radius changes from r 0 to r 0 + dr and the height does not change.
Solution: When we want to estimate the change in a quantity, we use differentials to do
so, where the differential of V relative to that of r is dV = dV
dV
dr
dr. Now
dr
= 1 3 π(2r)h
since only r is changing so that h is a constant. Hence a formula that estimates the
change in volume when the radius changes from r 0 to r 0 + dr (i.e., changes by the
quantity dr) is
dV = 2 3 πr 0hdr.
(b) Write a formula that estimates the change that occurs in the volume of a right circular
cone when the height changes from h 0 to h 0 + dh and the radius does not change.
Solution: As above, dV = dV
dh
dh estimates the change in volume since r is constant
and h varies in this part. Therefore,
for dV
dh = 1 3 πr2 .
dV = 1 3 πr2 dh,

Math 112, Calculus I
Test 2 Solutions
3. (25 points) Isocost lines and indifference curves are studied in economics. For what positive
value of C is the isocost line
5x + 45y = 1
tangent to the indifference curve
y = 1 x + C?
Find the point of tangency (called the equilibrium point).
Solution: We begin by finding where the indifference curve y = x −1 + C has the same slope
as that of the isocost line 5x + 45y = 1. The slopes are found by differentiating, as the
derivative measures the slope of a curve. Thus dy
dx = −x−2 gives the slope of the indifference
curve as a function of x while the slope of the isocost line is −1/9 (45y = −5x + 1 implies
that y = (−1/9)x + (1/45), so that the slope is −1/9). The curves have the same slope, then,
when −x −2 = −1/9, which happens when x 2 = 9. Thus the two curves are tangent when
x = 3 and when x = −3.
When x = 3, using the isocost line we find that y = (−1/9)[3] + (1/45) = −14/45, and
plugging these values in to the indifference curve we may find the corresponding value of C:
−14/45 = (1/3) + C when C = (−14/45) − (1/3), which is negative so it isn’t the value of C
we want.
When x = −3, we have y = (−1/9)[−3] + (1/45) = 16/45. Plugging this in to the indifference
curve, we find the corresponding value of C to be C = (16/45) − [1/(−3)] = 31/45, which is
positive, so is the value of C we are after. Hence the desired point of tangency is (−3, 16/45).
4. (25 points) A rubbish heap in the shape of a cube is being compacted. Given that the volume
decreases at the rate of 2 cubic inches per minute, find
(a) the rate of change of an edge and
(b) the rate of change of the total surface area
when the the volume of the cube is 64 cubic inches.
Solution: The volume of the cube is V = s 3 , where s refers to the length of one of its sides.
Differentiating this equation with respect to time, t, we have
dV
dt = [3s2 ] ds
dt .
Again, using V = s 3 , we find that s = 3√ V , so that s = √ 3[64] = 4 when V = 64. We are
told that dV
dt
= −2 (for the volume is decreasing), so we have −2 = 3(4) 2 ds ds
dt
= 48
dt
, so that
ds
dt = −2
48 = −1
24 .
The surface area of the cube, in terms of the length of one of its sides, is A = 6s 2 . Hence
dA
dt = 6[2s]ds dt = 12sds dt .
Therefore, as we found in part (a) that s = 4 and ds
dt = − 1
dA
dt = 12[4][− 1
24
] = −2 at this point in time.
24
when V = 64, we must have

Math 112, Calculus I
Test 2 Solutions
5. (30 points) Consider the function f(x) = x 3 − 2x 2 + x + 1.
(a) Find the intervals of increase or decrease.
Solution: To answer this question, we need to first find the derivative,
f ′ (x) = 3x 2 − 4x + 1 = (3x − 1)(x − 1),
together with its factorization as given above. Since this derivative is defined for every
real number x, the only critical numbers for f(x) occur when f ′ (x) = 0. Solving (3x −
1)(x − 1) = 0 we obtain x = 1/3 and x = 1. We therefore divide the real number line
into three intervals
(−∞, 1/3), (1/3, 1), (1, ∞)
and test points in each interval. The number 0 is in the first and f ′ (0) = 1 > 0 so f is
increasing on the entire interval (−∞, 1/3). The number 2/3 is in the middle interval,
(1/3, 1), and f ′ (2/3) = −1/3 < 0 implies that the function f is decreasing on the whole
interval (1/3, 1). Finally, at 2 in the interval (1, ∞) we have f ′ (2) = 5 > 0 so the function
f is increasing on this last interval.
(b) Find the local maximum and minimum values.
Solution: Using our answers in part (a) above, the only places where local extreme
values can occur are at the critical numbers 1/3 and 1. At x = 1/3, we found the
function was increasing to the left and decreasing to the right, so f has a local maximum
at the point (1/3, f(1/3)) = (1/3, 31/27). That is, the local maximum value of f is
f(1/3) = 31/27.
At x = 1, the function is decreasing to the left and increasing to the right, so we have a
local minimum at the point (1, f(1)) = (1, 1). Therefore, the local minimum value of f
is f(1) = 1.
(c) Find the intervals of concavity and the inflection points.
Solution: We do this problem in the same way as we did parts (a) and (b), but this
time using the second derivative for this is the function that tells us about the concavity
of f. We have already found f ′ (x) = 3x 2 − 4x + 1, so
f ′′ (x) = 6x − 4.
The function f ′′ (x) is always defined, so the concavity can only change where f ′′ (x) = 0,
i.e., when 6x − 4 = 0 at x = 2/3. The real number line is thus divided into two parts,
(−∞, 2/3) and (2/3, ∞). The number 0 is in the first interval where f ′′ (0) = −4 < 0 so
that the function is concave down on (−∞, 2/3). At the point 1 in the second interval,
we have f ′′ (1) = 2 > 0, from which we conclude that f is concave up on the interval
(2/3, ∞). Lastly, as we have just seen that the concavity of f indeed changes when
x = 2/3, the point (2/3, f(2/3)) = (2/3, 29/27) is the inflection point of the function f.

Math 112, Calculus I
Test 2 Solutions
6. (30 points) We are going to cut a rectangular beam of width w and depth d out of a 12-inch
diameter cylindrical log.
(a) Draw a picture of the cross-section of the cylindrical log (i.e., a circle) and how the
rectangular beam is cut out from it. Label the width w and depth d of the beam and
the diameter 12 of the log.
Solution: Your picture should be of a circle of diameter 12 in which you have drawn a
rectangle a rectangle each of whose vertices are points on the circle. The width of the
rectangle should be labelled by a w and its height by a d.
(b) How are the width w and the depth d related to one another?
Solution: If we draw the diameter of the circle that joins opposite vertices of the rectangle,
we find this line divides the rectangle into two right triangles, each of hypotenuse
12 and legs w,d. Therefore, the Pythagorean Theorem implies that
w 2 + d 2 = 12 2 = 144.
So we may write d 2 = 144 − w 2 and w 2 = 144 − d 2 . As w and d each measure lengths
within the circle, they must each take on values in [0, 12].
(c) The strength S of a rectangular wooden beam is proportional to its width w and the
square of its depth d, i.e., S = kwd 2 for some positive constant k. Find the dimensions
of the strongest beam that can be cut from a 12-inch cylindrical log.
Solution: From part (b), d 2 = 144 − w 2 , so that the strength of the beam is
S = kwd 2 = kw(144 − w 2 ) = k(144w − w 3 ).
In order to find the dimensions of the strongest beam, we seek to maximize the function
S above. We begin as always by finding its derivative, S ′ = k(144 − 3w 2 ). This is
always defined, so it only has a critical number where S ′ = k(144 − 3x 2 ) = 0. Since
k ≠ 0, we have S ′ = 0 only if 144 = 3w 2 , so w 2 = 144/3. Thus w = ± √ 144/3, of
which only the positive square root belongs to the domain of S, so it is our only critical
number. Considering the strengths for the endpoints w = 0 and w = 12 and the critical
number w = √ 144/3, we have S = 0, 0 and 665.1075k, respectively. Clearly this last
number is the largest, so is our absolute maximum, which occurs when w = √ √
144/3 and
d = 144 − ( √ 144/3) 2 = √ 288/3.