Math 160 - Assignment 8 - Summer 2012 - BSU - Jaimos F Skriletz
1. Area Between Curves
The area bounded by the curves y = 2x2 and y = 12 − x2 is as shown:
The intersection points are found by solving
2x2 = 12 − x2
3x2 = 12
x2 = 4
x = ±2
Thus the area is
Z 2
Z
2
2
[(12 − x ) − (2x )]dx =
−2
2
−2
2
[12 − 3x2 ]dx = [12x − x3 = [12(2) − (2)3 ] − [12(−2) − (−2)3 ] = 32
−2
1
2
Math 160 - Assignment 8 - Summer 2012 - BSU - Jaimos F Skriletz
2. Consumers’/Producers’ Surplus
Suppose that the supply and demand functions for a certain product are pS (x) = 14x + 60 and pD (x) = 300 − x2
respectively.
The equilibrium point occurs when
pS (x) = pD (x)
14x + 60 = 300 − x2
x2 + 14x − 240 = 0
(x + 24)(x − 10) = 0
x = −24
or
x = 10
Since sales must be positive the equilibrium demand is x = 10 and the equilibrium price is p = 200.
The graph of pS (x) and pD (x) is as shown
The Consumers’ Surplus is the blue area (as shown) and is
Z
10
Z
10
[pD (x)−p)dx =
0
2
Z
[(300−x )−200]dx =
0
0
10
10 1 3 1000
2000
[100−x ]dx = 100x − x = 1000 −
−[0] =
≈ $666.67
3 0
3
3
2
The Producers’ Surplus is the red area (as shown) and is
Z
10
Z
[p − pS (x)]dx =
0
10
Z
[200 − (14x + 60)]dx =
0
0
10
10
[140 − 14x]dx = [140x − 7x2 = [1400 − 700] − [0] = $700
0
3
Math 160 - Assignment 8 - Summer 2012 - BSU - Jaimos F Skriletz
3. Continuous Money Flow
Suppose that $12500 is invested at 4.3% interest compounded continuously. Further a continuous money flow of $50
per week is added to this investment.
(a) The future value of a continuous money flow at a rate of f (t) = $50/week = $2600/year is
e
rT
Z
T
f (t)e
−rt
dt = e
0
0.043T
Z
T
2600e
0
0.043T
=
2600e
0.043
−0.043t
dt = 2600e
0.043T
T
1
−0.043t e
−0.043
0
[−e−0.043T + e0 ] ≈ 60465.12[e0.043T − 1]
The future value of a $12,500 investment after T years is
P ert = 12500e0.043T
Thus the total value is the sum of the two investments
A(T ) = 12500e0.043T + 60465.12[e0.043T − 1] = 72965.12e0.043T − 60465.12
(b) The total after T = 5 years is
A(5) = 72965.12e0.043(5) − 60465.12 ≈ $30, 001.55
(c) If A(T ) = 50000 then
A(T ) = 72965.12e0.043T − 60465.12 = 50000
72965.12e0.043T = 110465.12
110465.12
≈ 1.514
e0.043T =
72965.12
0.043T = ln(1.514)
T =
Thus it will take about 9.6 years to reach a total of $50,000.
ln(1.514)
≈ 9.64
0.043
Math 160 - Assignment 8 - Summer 2012 - BSU - Jaimos F Skriletz
4
4. Partial Derivatives
Find all four second order partial derivatives of the two variable function f (x, y) = (y 3 − x3 )3
fx (x, y) = 3(y 3 − x3 )2 (−3x2 ) = −9x2 (y 3 − x3 )2
fxx (x, y) = [−18x](y 3 − x3 )2 + [2(y 3 − x3 )1 (−3x2 )](−9x2 ) = −18x(y 3 − x3 )[(y 3 − x3 ) − 3x3 ] = −18x(y 3 − x3 )(y 3 − 4x3 )
fxy (x, y) = −18x2 (y 3 − x3 )1 (3y 2 ) = −54x2 y 2 (y 3 − x3 )
fy (x, y) = 3(y 3 − x3 )2 (3y 2 ) = 9y 2 (y 3 − x3 )2
fyx (x, y) = 18y 2 (y 3 − x3 )1 (−3x2 ) = −54x2 y 2 (y 3 − x3 )
fyy (x, y) = [18y](y 3 − x3 )2 + [2(y 3 − x3 )1 (3y 2 )](9y 2 ) = 18y(y 3 − x3 )[(y 3 − x3 ) + 3y 2 ] = 18y(y 3 − x3 )(4y 3 − x3 )
5. Local Extrema of f (x, y)
Consider the function f (x, y) = 12xy − x3 − 2y 2
(a) The critical points are found by solving the system
fx (x, y) = 12y − 3x2 = 0
fy (x, y) = 12x − 4y = 0
The second equation gives us that 12x = 4y or y = 3x. Substitute this into the first equation to get
12(3x) − 3x2 = 0
3x(12 − x) = 0
x=0
or x = 12
If x = 0 then y = 3(0) = 0.
If x = 12 then y = 3(12) = 36.
Thus the two critical points are
(x1 , y1 ) = (0, 0)
and
(x2 , y2 ) = (12, 36)
(b) To determine what each critical point is use
D(x, y) = (fxx )(fyy ) − (fxy )2
For this function
fxx = −6x
fxy = 12
fyy = −4
D(x, y) = (−6x)(−4) − (12)2 = 24x − 144
For the critical point (x1 , y1 ) = (0, 0), D(0, 0) = 24(0) − 144 = −144 < 0. Thus f (0, 0) is a saddle point.
For the critical point (x2 , y2 ) = (12, 36), D(12, 36) = 24(12) − 144 = 144 > 0. Thus f (12, 36) is a local extrema.
Since fxx (12, 36) = −6(12) = −72 < 0 this extrema must be a maximum value. Thus f (12, 36) is a local maximum.