Math 108
Spring 2015
Show all your work
Dr. Lily Yen
Calculators allowed for this test.
Problem 1: Consider the following equation:
Test 3
Name:
Score:
/32
y 3 + 2x2 y − 8y = x3 + 19
a. Determine dy/dx.
Using implicit differentiation, 3y 2 y 0 + 4xy + 2x2 y 0 − 8y 0 = 3x2 , so
3y 2 y 0 + 2x2 y 0 − 8y 0 = 3x2 − 4xy, so (3y 2 + 2x2 − 8)y 0 = 3x2 − 4xy, so
y0 =
3x2 − 4xy
3y 2 + 2x2 − 8
Score:
/3
b. Find the equation of the tangent line at the point whose x-coordinate is 2.
If x = 2, then y 3 + 8y − 8y = 8 + 19, so y 3 = 27, so y = 3. Therefore the point of
tangency is (2, 3).
dy
12−24
= 27+8−8
= − 12
= − 49 . Thus the tangent line is
Again, if x = 2 and y = 3, then dx
27
given by
4
4
35
y − 3 = − (x − 2)
or
y =− x+
9
9
9
Score:
/2
Problem 2: The following parts are related.
a. State the Extreme Value Theorem.
If a (real-valued) function f is continuous in the closed (and bounded) interval [a, b],
then f must attain an absolute maximum and an absolute minimum, each at least
once.
Score:
/2
b. There are two conditions in the Extreme Value Theorem which together guarantee
the conclusion. Provide graphically a counter example for each of the two conditions
separately.
The interval [0, 1) is not closed
f is not continuous
Score:
/3
Problem 3: The number of salmon swimming upstream to spawn is approximated by
S(x) = −x3 + 3x2 + 260x + 5000,
for x ∈ [6, 20],
where x represents the temperature of the water in degrees Celsius. Find the water temperature accurate to 2 decimal places that produces the maximum number of salmon swimming
upstream. No marks for only numerical answers without appropriate calculus steps.
(
√
−8.363
Here S 0 (x) = −3x2 + 6x + 260, so S 0 (x) = 0 when x = 3± 3 789 ≈
. Of these only
10.36.
√
3+ 789
3
≈ 10.36 belongs to the interval [6, 20].
√
√
789
≈ 6904. Of these the last
Now, S(6) = 6452, S(20) = 3400, and S( 3+ 3 789 ) = 47358+526
9
is obviously the largest, so the (absolute) maximal number of salmon occurs when the
water temperature is 10.36 ◦C.
Score:
Page 2
/6
Math 108
√
Problem 4: Use the Linear Approximation Method to approximate 2 3 8.001 to 5 decimal
places. Compare your answer with your calculator’s evaluation. Show all your steps.
p
Let f be given by f (x) = 2 3 (x). Then f 0 (x) = 23 x−2/3 , and
√
2 3 8.001 = f (8.001) ≈ f (8) + f 0 (8) · 0.001 = 2 · 2 + 23 · 41 · 0.001 ≈ 4.000 166 666 67 which
compares rather well with the correct value of 4.000 166 659 72.
Score:
Page 3
/5
Math 108
Problem 5: A closed box with a square base is to have a volume of 16 000 cm3 . The material
for the top and bottom of the box costs $3.00 /cm2 , while the material for the sides costs
$1.50 /cm2 .
• Find the cost function.
• Find exact dimensions of the box that will lead to the minimum total cost.
• State the minimum total cost.
Let x be length of the side of the square base, and let h be the height. Then the volume is
. The cost is then
x2 h = 16000, so h = 16000
x2
C(x) = 3.00 · 2x2 + 1.50 · 4xh = 6x2 + 6x ·
16000
96000
,
= 6x2 +
2
x
x
where x > 0.
Therefore C 0 (x) = 12x − 96000
, so C 0 (x) = 0 when 12x3 = 96000, so x3 = 8000, so x = 20.
x2
192000
Now C 00 (x) = 12 + x3 , so C 00 (20) = 36 > 0, so the extremum at x = 20 is a minimum.
You could also use First Derivative Test to arrive at this conclusion.
The minimal cost is C(20) = $7200.
Score:
Page 4
/6
Math 108
Problem 6: A cone-shaped icicle is dripping from the roof. The radius of the icicle is
decreasing at a rate of 0.2 cm/hr, while the length is increasing at a rate of 0.8 cm/hr. If
the icicle is currently 4 cm in radius and 20 cm long, is the volume of the icicle increasing or
decreasing, and at what rate? Provide 2-decimal place accuracy.
= −0.2 cm/hr and dh
= +0.8 cm/hr when
Let r be the radius and h the length. Then dr
dt
dt
r = 4 cm and h = 20 cm.
= 23 πr dr
h + 31 πr2 dh
. Thus, when r = 4 cm and
Moreover, the volume is V = 13 πr2 h, so dV
dt
dt
dt
dV
2
1
2
h = 20 cm, dt = 3 π · 4(−0.2)20 + 3 π4 · 0.8 = −6.4π ≈ −20.11 cm3 /hr.
Hence the volume is decreasing at a rate of 20.11 cm3 /hr.
Score:
Page 5
/5
Math 108