MTH131 Applied Calculus – Spring 2016
Lab 5 – SOLUTIONS
1. Absolute extrema over a closed interval can occur at only two places: 1) at the endpoints of
the interval (in this case that’s x = −1 and x = 5), and 2) at any critical numbers that lie
within the interval. We had calculated the critical numbers in the previous lab.
(a) We first find the critical numbers for f (x) = x3 − 48x:
f 0 (x) = 3x2 − 48 = 0
3x2 = 48
x2 = 16
x = 4, −4
x = 4 is the only one critical number that lies in the interval, so we need to examine 3
points to determine the maximum and minimum of this function on the interval:
f (−1) = (−1)3 − 48(−1)
= 47 ⇐ maximum
f (4) = (4)3 − 48(4)
= −128 ⇐ minimum
f (5) = (5)3 − 48(5)
= −115
(b) We first find the critical numbers for f (x) = (2x − 8)2 :
f 0 (x) = 2(2x − 8)2 = 0
2x − 8 = 0
x = 4
x = 4 is lies in the interval, so we need to examine 3 points to determine the maximum
and minimum of this function on the interval:
f (−1) = (2(−1) − 8)2
= 100 ⇐ maximum
f (4) = (2(4) − 8)2
= 0 ⇐ minimum
f (5) = (2(5) − 8)2
= 4
(c) We first find the critical numbers for f (x) = (2x2 − 8)2 :
f 0 (x) = 2(2x2 − 8)4x = 0
(2x2 − 8)x = 0
(x2 − 4)x = 0
(x − 2)(x + 2)x = 0
x = 2, −2, 0
The critical numbers x = 0 and x = 2 both lie in the interval, so we need to examine 4
points to determine the maximum and minimum of this function on the interval:
f (−1) = (2(−1)2 − 8)2
= 36
f (0) = (2(0)2 − 8)2
= 64
f (2) = (2(2)2 − 8)2
= 0 ⇐ minimum
f (5) = (2(5)2 − 8)2
= 1764 ⇐ maximum
(d) We first find the critical numbers for f (x) = (3x + 6)2 (x − 2):
f 0 (x) = 2(3x + 6)(3)(x − 2) + (3x + 6)2 (1) = 0
6(3x + 6)(x − 2) + (3x + 6)2 = 0
(3x + 6)[6(x − 2) + (3x + 6)] = 0
(3x + 6)[6x − 12 + 3x + 6] = 0
(3x + 6)(9x − 6) = 0
x = 2/3, −2
x = 2/3 is the only critical number that lies in the interval, so we need to examine just
3 points to determine the maximum and minimum of this function on the interval:
f (−1) = (3(−1) + 6)2 ((−1) − 2)
= −27
f (2/3) = (3(2/3) + 6)2 ((2/3) − 2)
= −256/3 ⇐ minimum
f (5) = (3(5) + 6)2 ((5) − 2)
= 1323 ⇐ maximum
(e) Note that f (x) =
1
1
= 2
= (x2 − 4x − 5)−1 . We then find the
(x + 1)(x − 5)
x − 4x − 5
critical numbers:
f 0 (x) = −1(x2 − 4x − 5)−2 (2x − 4) = 0
2x − 4
−
= 0
(x2 − 4x − 5)2
−(2x − 4) = 0
x = 2
The function also has critical number at x = −1 and x = 5 as the first derivative is
undefined at those points. All three critical numbers x = −1, x = 2 and x = 5 lie in the
interval. The critical number x = 5 is a bit tricky, since the function is undefined there.
This means that the function either has no absolute maximum (if the function goes to
+∞ there) or no absolute minimum (if the function goes to −∞ there). Looking at the
two terms in the denominator we see that x + 1 is always positive on the interval [1, 5]
while the term x−5 is always negative. This means that on the interval the denominator
is always negative, so as x → 5, the function gets more an more negative, so at x = 5
the function value is −∞, so the function has no absolute minimum. This also applies
to the critical number x = −1.
To determine the absolute maximum we need to examine the remaining critical number
x = 2:
1
(2 + 1)(2 − 5)
1
= − ⇐ maximum
9
f (2) =
2. (a) Let x and y represent the various dimensions of the enclosure as shown below
x x
y
The area is given by A(x, y) = 2xy. Since there is only 1200 feet of fence, the values of x
and y are constrained by the equation 4x+3y = 1200 or y = (1200−4x)/3. Plugging this
into the area formula we obtain A(x) = 2x(1200 − 4x)/3 = (2/3)(1200x − 4x2 ). This is
maximized when the derivative A0 (x) = (2/3)(1200 − 8x) equals 0, or when x = 150 (and
thus y = (1200 − 4x)/3 = 200). Thus the area is maximized when the two enclosures
have dimensions 150 by 200, for a total area of A = 2(150)(200) = 60000 sq. feet.
(b) The enclosure would now look like the following
x
2x
y
The area is now A(x, y) = 3xy and the constraint equation is 6x+3y = 1200 or y = 400−
2x. Plugging this into the area formula we obtain A(x) = 3x(400 − 2x) = 1200x − 6x2 .
This is maximized when the derivative A0 (x) = 1200 − 12x equals 0, or when x = 100
(and thus y = 400 − 2x = 200). Thus the area is maximized when the two enclosures
have dimensions 100 by 200 and 200 by 200, for a total area of A = 3(100)(200) = 60000
sq. feet.
3. Let x be the dimension of the square cut out at each corner (see picture below)
x
12
x
16
Note that the maximum value for x would be 6 (since we are cutting squares from two ends of
the side of length 12), so the optimal value for x must lie in the range [0, 6]. The dimensions
of the box would then be 12 − 2x, 16 − x and x for a volume of V (x) = (12 − 2x)(16 − x)x =
192x − 44x2 + 2x3 . This is maximized when V 0 (x) = 0, or
192 − 88x + 6x2 = 0
96 − 44x + 3x2 = 0
(12 − x)(8 − 3x) = 0
x = 12, 8/3
The first value x = 12 is not in the allowed range, so the maximum is achieved when x = 8/3.
The resulting volume is V (8/3) = 237.04. Note that we should also check for the endpoints
of the interval [0, 6] but both of these points leads to a minimum volume of 0.
4. (a) The revenue for a given tax rate t is given by R(t) =(number of sales)×(tax rate per
sale) = S(t)t. For this problem this gives
R(t) = p
50t
.
(24t + 1)3
Plugging in the given values for t we obtain R(1) = 0.4, R(1/3) = 0.556, and R(0.1) =
0.798, each in units of millions of dollars.
(b) To find the tax rate that maximizes R(t), we find the derivative of this function and set
it equal to 0.
R0 (t) = 0
3
1
(24t + 1) 2 (50) − (50t)( 23 )(24t + 1) 2 (24)
(24t + 1)3
3
= 0,
1
50(24t + 1) 2 − 1800t(24t + 1) 2
(24t + 1)3
3
1
50(24t + 1) 2 − 1800t(24t + 1) 2
50(24t + 1)
3
2
= 0,
= 0,
1
= 1800t(24t + 1) 2 ,
50(24t + 1) = 1800t,
1200t + 50 = 1800t,
50 = 600t,
1
= t.
12
Thus the tax rate that maximizes the revenue is t = 1/12 = 0.08333 = 8.333%. The
revenue for this tax rate is R(1/12) = 0.802 million dollars, or $802,000.
5. Revenue R(x) = (quantity sold) × (unit price). Let x = number of $1 price increases. Then
the unit price will be 120 + x and the quantity sold will be 1000000 − 10000x. Thus the
revenue equation is
R(x) = (120 + x)(1000000 − 10000x) = 120000000 − 200000x − 10000x2
To find the maximum, we find where the derivative is 0
R0 (x) = −200000 − 20000x = 0
−20000x = 200000
x = −10
Thus to maximize revenue, the country should lower their price per barrel by $10 to $110.
They can expect to sell 1000000 − 10000(−10) = 1100000 barrels for a total revenue of
R(−10) = $121,000,000.
6. (a) With $5000 dollars, the most number of days she campaign is
5x2 + 500 = 5000
5x2 = 4500
x2 = 900
x = 30
Thus, the most votes she can gain is 2(30) = 60 thousand.
(b) In this problem, we simply need to find the value x which maximizes f (x) =
To do this, we set f 0 (x) equal to 0
(5x2 + 500)2 − 2x(10x)
(5x2 + 500)2
10x2 + 1000 − 20x2
(5x2 + 500)2
−10x2 + 1000
(5x2 + 500)2
2x
.
5x2 + 500
= 0
= 0
= 0
This is zero when the numerator −10x2 + 1000 = 0 or when x = 10. Thus the politician
should campaign for 10 days, gaining 2(10) = 20 thousand votes, spending only 5(102 ) +
500 = $1000. In this scenario, she gains (20000)/1000 = 20 votes per dollar; if she
spends all her money as in part (a), she only gains 60000/5000 = 12 votes per dollar.