Solutions for Calculus Midterm 2
(1) (i) Since the management sets a production goal of 20000 units, the constraint is
f(x, y) = 100x0.25y0.75= 20000.
That is
x0.25y0.75−200 = 0.
And we want to minimize the cost 48x+ 36y.
Then we use the method of Lagrange Multiplies and begin by writing the function
F(x, y, λ) = 48x+ 36y−λ(x0.25y0.75−200).
Set
Fx = 48−0.25λ(y
x)0.75= 0.
Fy = 36−0.75λ(x
y)0.25= 0.
Fλ = 200−x0.25y0.75= 0.
This implies y x = 4.
Thus x= 50√
2 and y= 200√ 2.
(ii)
Marginal productivity of labor Marginal productivity of capital
= fx(x, y) fy(x, y)
= 25x−0.75y0.75 75x0.25y−0.25
= 1 3(y
x)
= 4
3 by (i)
= 48 36
= unit price of labor unit price of capital.
1
(2) (i) The sum of squared errors of y = ax +b with respect to the n points is
S = [(ax1+b)−y1]2+ [(ax2+b)−y2]2+...+ [(axn+b)−yn]2
=
n
X
i=1
(axi+b−yi)2.
(ii)
A= nΣxy−(Σx)(Σy) nΣx2−(Σx)2
= 4×91−13×21 4×51−169
= 2.6 B = 1
n(Σy−AΣx)
= 1
4(21−2.6×13)
=−3.2
Thus the least squares regression line is y= 2.6x−3.2.
(3) (i) We show it in Figure 1.
R
y=x y=2x
0 1 2 3 4 5 6
0.5 1 1.5 2 2.5 3
x
Figure 1: The regionR onx−y plane.
(ii)
Z Z
R
y
x2+y2dA = Z 2
0
Z 2x
x
y
x2+y2dydx Z Z
R
y
x2+y2dA = Z 2
0
Z y
y 2
y
x2 +y2dxdy+ Z 4
2
Z 2
y 2
y
x2+y2dxdy
3
(iii)
Z 2
0
Z 2x
x
y
x2+y2dydx= Z 2
0
(1
2ln(x2 +y2)
y=2x
y=x
)dx
= Z 2
0
1
2(ln 5x2−ln 2x2)dx
= Z 2
0
1 2ln5x2
2x2dx
= 1 2
Z 2
0
ln5 2dx
= 1 2ln5
2x
2
0
= ln5 2 (4) (i) f(t) = sin(2πt
24 ) is a periodic function with period 24.
h(t) = (sint)2 = 1−cos 2t
2 is also a periodic function with period π.
(ii) We sketch those graphs in Figure 2 and Figure 3.
–1 –0.5
0 0.5 1
–20 –10 10 20
t
Figure 2: f(t) = sin(2πt 24).
0 0.2 0.4 0.6 0.8 1
–6 –4 –2 2 4 6
t
Figure 3: h(t) = sin2t. 5
(5) F(t) attains minimum when sin2π(t−30)
365 =−1.
⇒ 2π(t−30) 365 = 3π
2 .
⇒t= 303.75≈304.
Thus the minimum sales occur on the 304st day of the year, that is on October 31.
(6)
xlim→∞
x
ex = lim
x→∞
1 ex = 0.
xlim→∞
x0.001
x = lim
x→∞
0.001x−0.999
1 = 0.
xlim→∞
lnx
x0.001 = lim
x→∞
1 x
0.001x−0.999 = 0.
xlim→∞
ln(lnx)
lnx = lim
x→∞
1 lnx
1 x 1 x
= lim
x→∞
1 lnx = 0.
Thus we order them asg(x) = ex,f(x) =x, I(x) =x0.001,h(x) = lnx, J(x) = ln(lnx).
(i) g(x) =ex grows to infinity with the most rapid rate.
(ii) J(x) = ln(lnx) grows to infinity with the slowest rate.
(iii) The middle isI(x) = x0.001. (7) (i)
Z 1
0
tan(1−x)dx= Z 1
0
sin(1−x) cos(1−x)dx
= ln|cos(1−x)|10
= ln 1−ln(cos 1)
=−ln(cos 1)
(ii) Let u = x, dv = cosxdx. Then du = dx and v = sinx. Apply integration by parts and obtain
Z
xcosxdx=xsinx− Z
sinxdx
=xsinx+ cosx+C.