Integrated Calculus I Final Practice Solutions
Question 1
Find the derivative of each of the following functions:
(a) f (t) = sin(t)et
f 0 (t) = cos(t)et + sin(t)et .
Here we used the product rule (uv)0 = u0 v + uv 0 and the derivatives:
d
sin(u) = cos(u) du
and dtd eu = eu du
.
dt
dt
dt
(b) g(x) =
tan(x)
ln(x)
g 0 (x) =
sec2 (x) ln(x) − tan(x)( x1 )
.
(ln(x))2
Here we used the division rule ( uv )0 =
d
d
tan(u) = sec( u) du
and dx
ln(u) = u1 du
.
dx
dx
dx
u0 v−uv 0
v2
and the derivatives:
√
(c) y(x) = x1+ x
We first take logarithms of both sides of the equation, using ln(ab ) =
b ln(a) and then differentiate, using the product rule and the power rule
d n
u = nun−1 du
:
dx
dx
√
ln(y) = (1 + x) ln(x),
√ 1
1 0 1 −1
y = x 2 ln(x) + (1 + x) ,
y
2
x
√ 1
1 1
y 0 = y( x− 2 ln(x) + (1 + x) ),
2
x
√ 1
√
1
1
y 0 = x1+ x ( x− 2 ln(x) + (1 + x) ).
2
x
(d) z(t) = cos(1 + ln(t))
We use the chain rule and the derivative:
d
dx
cos(u) = − sin(u) du
:
dx
1
z 0 (t) = − sin(1 + ln(t))( ).
t
Question 2
Calculate the following indefinite integrals:
Z
(a) (6x + cos(x) − 4 sin(2x))dx
We need F (x), such that F 0 (x) = 6x + cos(x) − 4 sin(2x).
We first find functions a(x), b(x) and c(x), whose derivatives obey:
a0 (x) = 6x, b0 (x) = cos(x) and c0 (x) = −4 sin(2x).
d
d
– We have dx
(x2 ) = 2x, so dx
(3x2 ) = 6x, so we may take a(x) = 3x2 .
d
– We know that dx
(sin(x)) = cos(x), so we may take b(x) = sin(x).
d
– We have dx (cos(2x)) = −2 sin(2x), so we may take c(x) = 2 cos(2x).
Then F (x) = a(x) + b(x) + c(x) obeys F 0 (x) = 6x + cos(x) − 4 sin(2x),
so by the definition of the indefinite integral, we have:
Z
(6x+cos(x)−4 sin(2x))dx = F (x)+C = 3x2 +cos(x)+2 cos(2x)+C.
Z
(b)
3
dx
2 + 2x2
3
3
1
We need G(x), such that G0 (x) = 2+2x
.
2 = 2
1+x2
d
1
3
We have dx arctan(x) = 1+x2 , so G(x) = 2 arctan(x) will do, giving:
Z
3
3
dx = G(x) + C = arctan(x) + C.
2
2 + 2x
2
Question 3
Evaluate the following integrals, showing your work:
Z 2
1
(a)
(4 + t2 + )dt
t
1
We first need F (t), such that F 0 (t) = 4 + t2 + 1t .
3
Clearly we may take F (t) = 4t + t3 + ln(t). (Here ln(t) is allowed,
rather than ln(At), since we only need the case that t > 0 for the
definite integral). Then by FTC, the answer to the integral is:
Z 2
1
(4 + t2 + )dt = [F (t)]21 = F (2) − F (1)
t
1
= 4(2) +
23
13
19
+ ln(2) − (4(1) +
+ ln(1)) =
+ ln(2) = 7.026480514.
3
3
3
2
Z
1
2
tet dt
(b)
0
2
– We first need G(t), such that G0 (t) = tet .
2
2
Now dtd et = 2tet , which is twice what we want, so we may take
2
G(t) = 21 et .
– Then by FTC, the answer to the integral is:
Z 1
1
2
tet dt = [G(t)]10 = G(1) − G(0) = (e − 1) = 0.8591409.
2
0
Z
π
4
(c)
(cos(t) − sec2 (t))dt
0
– We first need H(t), such that H 0 (t) = cos(t) − sec2 (t).
Clearly we may take H(t) = sin(t) − tan(t).
– Then by FTC, the answer to the integral is:
Z
0
π
4
π
π
(cos(t) − sec2 (t))dt = [H(t)]04 = H( ) − H(0)
4
π
π
1
= sin( )−tan( )−(sin(0)−tan(0)) = √ −1 = −0.2928932188.
4
4
2
Z
(d)
1
e−3t dt
0
– We first need J(t), such that J 0 (t) = e−3t .
Now dtd e−3t = −3e−3t , which is (−3) times what we want, so we
may take J(t) = − 13 e−3t .
– Then by FTC, the answer to the integral is:
Z 1
1
e−3t dt = [J(t)]10 = J(1) − J(0) = − (e−3 − 1) = 0.316737644.
3
0
3
Question 4
Find the area bounded by the curve y = sin(x) + cos(x), the x-axis and the
lines x = 0 and x = π.
• Plotting the region of interest, we see that the curve y = sin(x)+cos(x)
crosses the x-axis at x = 3π
, going from positive to negative, when
4
sin(x) = √12 and cos(x) = − √12 .
• So the area in question is given by A = J1 − J2 , where we have:
R 3π
Rπ
J1 = 0 4 (sin(x) + cos(x))dx and J2 = 3π (sin(x) + cos(x))dx.
4
0
• Let F (x) = sin(x) + cos(x).
Then by FTC, we have J1 = F ( 3π
) − F (0) and J2 = F (π) − F ( 3π
), so
4
4
3π
3π
3π
we get: A = J1 − J2 = F ( 4 ) − F (0) − (F (π) − F ( 4 )) = 2F ( 4 ) −
F (0) − F (π).
• Clearly we may take F (x) = − cos(x) + sin(x), giving:
A = 2F (
= −2 cos(
3π
) − F (0) − F (π)
4
3π
3π
) + 2 sin( ) − (− cos(0) + sin(0)) − (− cos(π) + sin(π))
4
4
√
√
√
= 2 + 2 + 1 − 1 = 2 2 = 2.82842712474619.
Question 5
Evaluate each of the following limits or explain why the limit does not exist.
x−3
x→3 |x − 3|
(a) lim
x−3
|x−3|
– If we put x = 3 + t, where t > 0, then
x−3
limx→3+ |x−3|
= 1.
– If we put x = 3 + t, where t < 0, then
x−3
limx→3− |x−3|
= −1.
x−3
|x−3|
x−3
– Since limx→3+ |x−3|
does not equal limx→3−
does not exist.
4
=
=
x−3
,
|x−3|
t
|t|
t
|t|
=
=
t
t
= 1, so
t
−t
= 1, so
the required limit
x2 − x − 12
x→−3
x+3
We factor, remove the bad factor x + 3 and take the limit:
(b) lim
x2 − x − 12
(x − 4)(x + 3)
= lim
x→−3
x→−3
x+3
x+3
lim
= lim (x − 4) = −3 − 4 = −7.
x→−3
3n2 + 2n − 4
(c) lim √
. For large n we have approximately
n→∞
n4 − n3 + 1
so the required limit is 3.
More formally we may write the calculation as follows:
3n2
√
n4
=
3n2
n2
= 3,
n2 (3 + n2 − n42 )
n2 (3 + n2 − n42 )
3n2 + 2n − 4
lim √
= lim q
= lim q
n→∞
n4 − n3 + 1 n→∞ n4 (1 − 1 + 14 ) n→∞ n2 (1 − 1 + 14 )
n
n
n
n
(3 + n2 − n42 )
(3 + 0 − 0)
=√
= 3.
= lim q
n→∞
1
1
1
−
0
+
0
(1 − n + n4 )
x sin(x) − cos(x) + 1
(d) lim
x→0
x2
This is a ”zero over zero” limit, so we may use L’Hopital’s rule:
x sin(x) − cos(x) + 1
= lim
x→0
x→0
x2
lim
d
(x sin(x) − cos(x)
dx
d
(x2 )
dx
+ 1)
sin(x) + x cos(x) + sin(x)
2 sin(x) + x cos(x)
= lim
.
x→0
2x
2x
Again, this is a ”zero over zero” limit, so we may use L’Hopital’s rule
again:
= lim
x→0
2 sin(x) + x cos(x)
= lim
x→0
x→0
2x
lim
d
(2 sin(x) + x cos(x))
dx
d
(2x)
dx
2 cos(x) + cos(x) − x sin(x))
x→0
2
1
3
= (2 cos(0) + cos(0) − 0) = .
2
2
= lim
5
Question 6
A curve is given implicitly by the following equation:
−xy + x3 + cos(y) = 9.
• Find the equations of the tangent and normal lines through the point
(x, y) = (2, 0) on the curve.
Differentiating the equation of the curve, thinking of y as a function of
x gives:
dy
dy
= 0.
−y − x + 3x2 − sin(y)
dx
dx
Putting x = 2 and y = 0 in this equation gives:
0−2
dy
+ 12 − 0 = 0,
dx
dy
= 6.
dx
So the tangent line has slope 6 at the given point, so the equation of
the tangent line is: y − 0 = 6(x − 2), or y = 6x − 12 and the equation
of the normal line is y − 0 = − 61 (x − 2), or x + 6y = 2.
• Using an appropriate linear approximation, or otherwise, estimate the
value of y on the curve when x = 2.1.
By the previous part the linear approximation based at x = 2 is y =
6x − 12.
21
Putting x = 2.1 = 10
gives y approximately as 6( 21
) − 12 = 35 =
10
0.6. Maple gives a more accurate value for y as 0.534, so the linear
approximation is not too bad.
• If a particle moves along the curve in such a way that dx
= 3 at the
dt
point (2, 0), find its velocity vector and speed at that point.
( dy )
( dy )
dy
dt
= dt3 , using the chain rule, so dy
= 18.
We have 6 = dx
= ( dx
dt
)
dt
Alternatively we differentiate the equation of the curve with respect to
time:
−x0 y − xy 0 + 3x2 x0 − sin(y)y 0 = 0.
Putting x0 = 3, x = 2 and y = 0 in this equation gives: 0 − 2y 0 +
3(22 )(3) − 0 = 0, or 2y 0 = 36, so y 0 = 18, as before.
, dy ] = [6, 18] and the speed is
Then the velocity vector is: V = [ dx
√ dt dt
√
√
|V | = 62 + 182 = 6 1 + 32 = 6 10 = 18.973665961.
6
Question 7
A ten kilogram mass is hanging in a room at the point B = [4, 4].
It attached by straight strings to points A = [0, 8] and C = [12, 8] on a ceiling
above the mass.
Find the forces in the strings AB and BC, assuming that the strings have
negligible mass.
• The direction vector along the line BA is the vector:
BA = A − B = [0, 8] − [4, 4] = [−4, 4].
So the force F along BA may be written F = sBA = s[−4, 4] =
[−4s, 4s], for some number s.
• The direction vector along the line BC is the vector:
CB = C − B = [12, 8] − [4, 4] = [8, 4].
So the force G along BC may be written G = t[8, 4] = [8t, 4t], for some
number t.
• The force of gravity on the mass is H = [0, −10g].
• Since the mass is not moving, the total force acting on it is zero, giving
the vector equation:
[0, 0] = F +G+H = [−4s, 4s]+[8t, 4t]+[0, −10g] = [−4s+8t, 4s+4t−10g],
−4s + 8t = 0,
4s + 4t − 10g = 0,
s = 2t, 8t + 4t − 10g = 0,
5g
5g
t= , s= .
6
3
, 20g
]=
• So the force in the string
AB is F = [−4s, 4s] = [− 20g
3
3
√
20 2g
[−65.4, 65.4] of size 3 = 92.489567 Newtons.
20g
[−1, 1]
3
, 10g
]=
• Finally the force in the√string BC is G = [8t, 4t] = [ 20g
3
3
10 5g
[−65.4, 32.7] of size 3 = 73.119423 Newtons.
10g
[2, 1]
3
7
=
=
Question 8
1 kilogram of a dangerous radioactive chemical is delivered to a laboratory
for an experiment. After 3 days, 800 grams of the chemical remains.
The experiment needs at least 600 grams of the chemical to start successfully.
When is the latest time after delivery that the experiment can be successfully
started? The experiment will end when 1 gram of the material remains.
When will that be?
• After three days the amount of material has been multiplied by a factor
800
of 1000
= 45 , so the amount A(t) grams at time t after delivery is given
t
by the formula: A(t) = 1000( 45 ) 3 . Then we have A(t) = 600, when
t
3 ln( 3 )
t
600
600 = 1000( 45 ) 3 , or 1000
= 35 = ( 45 ) 3 , or 3t ln( 45 ) = ln( 35 ), or t = ln( 45 =
5
6.867673 days or after about six days, twenty hours, forty-nine minutes
and twenty-seven seconds.
t
t
1
Finally, we have A(t) = 1, when 1 = 1000( 45 ) 3 , or 1000
= ( 45 ) 3 , or
1
3 ln(
t
3
)
1
ln( 45 ) = ln( 1000
), or t = ln(1000
= 92.86966 days or after about
4
5
ninety-two days, twenty hours, fifty-two minutes and nineteen seconds.
• Alternatively, we use the half-life formalism: we first write A(t) =
t
1000( 21 ) T , where T is the half-life.
3
3
Then at t = 3, we have 800 = 1000( 21 ) T , so 45 = ( 12 ) T , so ln( 45 ) =
3
T
ln( 12 ), so T =
3 ln( 21 )
ln( 45 )
= 9.318851 days.
t
Then we have A(t) = 600, when 600 = 1000( 21 ) T , so
ln( 53 ) =
t
T
ln( 12 ), so t =
T ln( 35 )
ln( 12 )
ln( 12 ), so t =
1
)
T ln( 1000
1
ln( 2 )
t
= ( 12 ) T , so
= 6.867673 days, as before. Finally,
t
we have A(t) = 1, when 1 = 1000( 12 ) T , so
t
T
3
5
1
1000
t
1
= ( 12 ) T , so ln( 1000
)=
= 92.86966 days, as before.
• Alternatively, we use the exponential formalism: we first write A(t) =
1000e−kt , where k is the decay constant.
Then at t = 3, we have 800 = 1000e−3k , so 45 = e−3k , so ln( 45 ) = −3k,
so k = −
ln( 54 )
3
so 53
= 0.0743812 per day. Then A(t) = 600, when 600 =
ln( 3 )
1000e−kt ,
= e−kt , so ln( 35 ) = −kt, so t = − k5 = 6.867673 days,
1
as before. Finally, we have A(t) = 1, when 1 = 1000e−kt , so 1000
= e−kt ,
1
so ln( 1000
) = −kt, so t = −
1
ln( 1000
)
k
8
= 92.86966 days, as before.
Question 9
A child drags her 10 kilogram sled 30 meters smoothly up a snow covered
slope.
The slope makes an angle of 12 degrees with the horizontal.
The child applies a constant force F of size 150 Newtons to the sled, at an
angle of 15 degrees to the slope.
• What is the force G due to gravity on the sled (the acceleration due to
gravity is 9.81 meters per second per second)?
We have G = [0, −10g] = [0, −98.1].
• What is the resultant force H of gravity and the child, acting on the
sled? How large is the force H?
The force F makes a total angle of 15+12 = 27 degrees, or θ =
radians with the horizontal, giving:
F = [150 cos(θ), 150 sin(θ)] = [150 cos(
27π
180
=
3π
20
3π
3π
), 150 sin( )] = [133.65, 68.10].
20
20
Then the resultant force asked for is:
3π
3π
), 150 sin( ) − 10g] = [133.65, −30.00].
20
20
p
The size of the resultant is |H| = (133.65)2 + (−30)2 = 136.98 Newtons.
H = F + G = [150 cos(
• How much work is done by the child on the sled in dragging the sled
the 30 meters?
The component of the force F in the direction of motion is 150 cos(φ),
π
= 12
radians.
where φ = 15 degrees, or φ = 15π
180
π
Then the work done by the child is 150 cos(φ)(30) = 4500 cos( 12
) =
4401.7 Joules.
9
Question 10
Consider the function y = t3 − 6t2 − 15t + 20 on the interval [−5, 10].
• Find the first and second derivatives of the function y.
We have y 0 = 3t2 − 12t − 15 = 3(t2 − 4t − 5) = 3(t + 1)(t − 5) and
y 00 = 6t − 12 = 6(t − 2).
• Find the critical points of the function y.
We need y 0 = 0, so (t + 1)(t − 5) = 0, so t = −1 or t = 5.
When t = −1, we have y = −1 − 6 + 15 + 20 = 28 and when t = 5, we
have y = 125 − 150 − 75 + 20 = −80.
So the critical points are (−1, 28) and (5, −80).
• Find the intervals on which the function y is increasing and the intervals
on which the function y is decreasing.
y is increasing, when y 0 ≥ 0, so when the factors (t + 1) and (t − 5) have
the same sign, so when t ≥ 5, or t ≤ −1, so on the intervals [5, ∞) and
(−∞, −1]. y is increasing, when y 0 ≤ 0, so when the factors (t + 1) and
(t − 5) have the opposite sign, so when t ≤ 5 and t ≥ −1, so on the
interval [−1, 5].
• Find the local maxima and minima of the function y.
At t = −1, we have y 00 < 0, so t = −1 is a local maximum.
At t = 5, we have y 00 > 0, so t = 5 is a local minimum.
So (−1, 28) is the only local maximum and (5, −80) is the only local
minimum.
• On what intervals is the graph of the function y concave down?
The graph is concave down, when y 00 ≤ 0, so when t ≤ 2, so on the
interval (−∞, 2].
• Sketch the graph of the function y.
10
Question 11
For x any real number, define:
x
Z
C(x) =
t sin(t)dt.
0
• What is the derivative of C(x)?
By FTC, we have: C 0 (x) = x sin(x).
• For which intervals in x is the function C(x) increasing?
Explain your answer.
We need C 0 (x) ≥ 0, so when x ≥ 0, we need sin(x) ≥ 0, so x lies in the
intervals [2mπ, (2m + 1)π], where m ≥ 0 is an integer and when x ≤ 0,
we need sin(x) ≤ 0, so x lies in the intervals [−(2n + 1)π, −2nπ], where
n ≥ 0 is an integer.
• What is the smallest positive x0 , such that x = x0 is a local maximum
of the function C(x)?
Explain your answer.
We try the first positive x where C 0 (x) = x sin(x) = 0.
This gives x = π.
We have C 00 (x) = sin(x) + x cos(x), so C 00 (π) = sin(π) + π cos(π) = −π.
Since C 00 (π) < 0, the curve y = C(x) is concave down at x = π, so
x = π is a local maximum and so x0 = π.
11
Question 12
Estimate numerically the area under the graph of the function y = x2 + x,
for x = 0 to x = 5:
• Estimate L: use 5 equal intervals and left end-points.
Let f (x) = x2 + x, for any real x.
We have ·§ = 5−0
= 1, so our data points are spaced at intervals of one
5
unit.
L = 1(f (0) + f (1) + f (2) + f (3) + f (4)) = 0 + 2 + 6 + 12 + 20 = 40.
• Estimate R: use 5 equal intervals and right end-points.
R = 1(f (1) + f (2) + f (3) + f (4) + f (5)) = 2 + 6 + 12 + 20 + 30 = 70.
• Estimate M: use 5 equal intervals and the mid-points of each interval.
1
3
5
7
9
3 15 35 63 99
215
M = 1(f ( )+f ( )+f ( )+f ( )+f ( )) = + + + + =
= 53.75.
2
2
2
2
2
4 4 4 4 4
4
• Also sketch your approximating rectangles and the graph of the function y on one graph.
• How do your estimates relate to the actual area under the graph?
Explain your answer.
The function is increasing, f 0 (x) = 2x+1 > 0, and concave up, f 00 (x) =
2 > 0, so each of the rectangles of L lies below the curve and each of
the rectangles of R lies above the curve. Also L is the lower Riemann
sum and R is the upper Riemann sum. Each rectangle of M may
be replaced by a trapezoid of equal area tangent to the curve at the
point above the midpoint of the interval in question. Since the curve
is concave up, these trapezoids lie below the curve.
So if A is theRtrue area, we have L < M < A < R.
3
2
3
2
5
In fact A = 0 (x2 + x)dx = [ x3 + x2 ]50 = 53 + 52 = 325
= 54.166, in
6
agreement with these inequalities.
• Why is the estimate L+R
a better estimate than either L or R?
2
Explain your answer graphically.
The estimate 12 (L + R) = 12 (40 + 70) = 55 is the trapezoidal rule: its
trapezoids are much closer to the graph than are the rectangles of either
the rules L or R, so it gives a better estimate.
12
Question 13
4
Let f (x) = x + , defined for any x 6= 0.
x
• From first principles, compute the following limits:
f (x) − f (1)
.
x→1
x−1
f (x) − f (2)
L2 = lim
.
x→2
x−2
We have f (1) = 1 + 4 = 5 and f (2) = 2 + 2 = 4, giving:
L1 = lim
x + x4 − 5
x2 + 4 − 5x
= lim
x→1
x→1 x(x − 1)
x−1
L1 = lim
(x − 4)(x − 1)
x−4
1−4
= lim
=
= −3,
x→1
x(x − 1)
x
1
x + x4 − 4
x2 + 4 − 4x
= lim
L2 = lim
x→2 x(x − 2)
x→2
x−2
x−2
2−2
(x − 2)(x − 2)
= lim
= lim
=
= 0.
x→2
x→2
x(x − 2)
x
2
= lim
x→1
• Give the geometrical interpretation of the limits L1 and L2 .
L1 gives the slope of the tangent line to the curve y = x +
point (x, y) = (1, 5).
L2 gives the slope of the tangent line to the curve y = x +
point (x, y) = (2, 4).
4
x
at the
4
x
at the
• Find the equations of the tangent and normal lines to the curve y =
f (x) at x = 1 and at x = 2.
– The tangent line at (1, 5) has the equation y − 5 = −3(x − 1), or
y + 3x = 8.
– The normal line at (1, 5) has the equation y − 5 = − 31 (x − 1), or
x − 3y = −14.
– The tangent line at (2, 4) is horizontal, so has the equation y = 4.
– The normal line at (2, 4) is vertical, so has the equation x = 2.
• Sketch the curve and your tangent and normal lines.
13
Question 14
A particle has position vector X at time t seconds, given by the formula:
X = [sin(t) − cos(2t), cos(t) − sin(2t)].
Find its velocity, acceleration and speed.
• The velocity is:
V =
dX
= [cos(t) + 2 sin(2t), − sin(t) − 2 cos(2t)]
dt
• The acceleration is:
A=
dV
= [− sin(t) + 4 cos(2t), cos(t) + 4 sin(2t)]
dt
• Then we have:
|V |2 = (cos(t) + 2 sin(2t))2 + (− sin(t) − 2 cos(2t))2
= cos2 (t)+4 sin2 (2t)+4 sin(2t) cos(t)+sin(t)2 +4 cos2 (2t)+4 sin(t) cos(2t)
= 5 + 4 sin(3t).
Here we used the trigonometric identities cos2 (t) + sin2 (t) = 1 and
sin(a + b) = sin(a) cos(b)
p + cos(a) sin(b).
So the speed is |V = 5 + 4 sin(3t).
Show that |V |2 + 2|X|2 is constant and compute that constant.
We have:
|X|2 = (sin(t) − cos(2t))2 + (cos(t) − sin(2t))2
= sin2 (t) + cos2 (2t) − 2 sin(t) cos(2t) + cos2 (t) + sin2 (2t) − 2 cos(t) sin(2t)
= 2 − 2 sin(3t),
|V |2 + 2|X|2 = 5 + 4 sin(3t) + 2(2 − sin(3t)) = 9.
14
Find its minimum and maximum speeds and the times at which these minimum and maximum speeds occur.
Show that the minimum speed occurs when the particle is furthest from the
origin and the maximum speed occurs when the particle is at the origin.
• The minimum speed of 1 occurs when sin(3t) = −1, so when t =
− π6 + 2nπ
or t = π2 + 2mπ
, where m and n are arbitrary integers and
3
3
p
then since |X| = 2 − 2 sin(3t)|, |X| reaches its maximum value of 2.
• The maximum speed of 3 occurs when sin(3t) = 1, so when t = π6 + 2pπ
3
or t = − π2 + 2qπ
,
where
p
and
q
are
arbitrary
integers
and
then
since
3
p
|X| = 2 − 2 sin(3t)|, |X| reaches its minimum value of 0: i.e. the
particle is at the origin.
Question 15
A fence is to be made to enclose 5000 square feet of a field, in the shape of
a rectangle ABCD.
The material for the two parallel sides AB and CD of the fence costs twice
as much as the material for the other two parallel sides BC and DA.
What should be the dimensions of the fence in order to minimize the total
cost of materials?
Let the length of the expensive sides be x feet and of the cheaper sides be y.
Then the total cost is represented by the function C = 4x + 2y.
The area of the field is xy = 5000, so y = 5000
.
x
10000
Then C = 4x + 2y = 4x + x .
At a local minimum for the cost, we have C 0 = 0, so 0 = C 0 (x) = 4 − 10000
,
x2
so x2 = 2500, so x = 50 (since x > 0, for this problem).
4
We have C 00 (50) = 20000
= 25
> 0, so x = 50 does give a local minimum.
503
+
As x → 0 and as x → ∞, we have C(x) → ∞, so there are no other
extrema, so x = 50 gives the desired minimum cost.
When x = 50 feet, we have y = 100 feet.
So the two more expensive sides should each be 50 feet long and the two
cheaper sides should each be 100 feet long.
15