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course MTH 173
12/17 17:40
Text Section 4.5Text Problem 4.5.1
Given curves are cost curves and the revenue curves. The profit function pi(q) = R(q) - C(q)
The profit function will be positive when R(q) - C(q) is positive, that is when R(q) > C(q), that is when R(q) function lies above C(q) function. R(q) function lies above the C(q) function for q values from 5.5 to 12.5. At q = 5.5 and 12.5, the revenue and the cost is equal and thus profit is zero. For 5.5 < q < 12.5, since R(q) function is above C(q) function, the profit is positive. For q values greater than 12.5 and less than 5.5, the cost function is greater than revenue function and thus the profit is negative. The value of the profit is maximum when R’(q) = C’(q) in the interval of (5.5,12.5). This is equivalent to to say the q value at which the vertical distance between the revenue curve and the cost curve is greatest provided the revenue function is above the cost function. In this curve this occurs when q = approximately equal to 9.5. Thus the function will have maximum profit when q = 9.5 thousand
Text Problem 4.5.3
Given cost function
C(q) = 5000 + 2.4q
And the revenue function
R(q) = 4q
Fixed cost is defined as the value of the cost function C(q) when q = 0
C(0) = 5000 + 2.4*0 = $5000
Marginal cost per item will be defined as the value of the marginal cost
Marginal cost = MC = C’(q)
Using the rules of differentiation C’(q) = (5000 + 2.4q)’ = 2.4
$2.4 is the marginal cost.
The price at which this quantity is sold is the marginal revenue generated which is
MR = R’(q) = (4q)’ = $4
Thus marginal cost per item is $2.4 and the cost at which this quantity is sold is $4.
Text Problem 4.5.6
Given the initial cost incurred before cleaning any house to the couple is $5,000.
Let q be the number of house cleaned.
Given that $15 is the cost of supplies to the couple for cleaning 1 house.
Thus the cost of supplies for cleaning q houses will be 15q
Thus the total cost function C(q) = $5000 + $15q
Given that they earn $60 for 1 house cleaned.
Thus the revenue earned for cleaning q house is
R(q) = $60 * q = $60q.
The profit function would thus be the difference of the revenue and the cost function
pi(q) = R(q) - C(q)
pi(q) = $60q - $5,000 - $15q
pi(q) = $45q - $5,000
Text Problem 4.5.8
Given revenue function R(q) = 500q - q^2
And the cost function is C(q) = 150 + 10q
The total profit earned will be the difference of the revenue and the cost function
Thus the total profit earned function will be
pi(q) = R(q) - C(q) = 500q - q^2 - 150 - 10q = 490q - q^2 - 150
the profit will be maximum when pi’(q) would be 0
Using the rules of differentiation we get
pi’(q) = 490 - 2q = 0
q = 490/2 = 245 units.
R(245) = 500*245 - 245*245 = 245*255 = $62,475
C(245) = 150 + 10*245 = 2,450 + 150 = $2,600
Since the R(245) Is greater than C(245) , q = 245 units is the quantity that maximizes the profit and the value of this profit is pi(245) = $62,475 - $2,600 = $59,875
Text Problem 4.5.12
Given a table of Marginal cost and Marginal revenue
(a)
To determine whether production should be increased or decreased from 5000 we need to calculate the MR and MC at q = 5000
From the table MR = 60 and MC = 48.
Since MR > MC extra revenue earned on increasing q would be greater than the extra cost incurred on adding q, thus there is a profit on adding q that is increasing q from 5000
Thus production should be increased from q = 5000.
(b)
We know that the profit is maximum when MR = MC.
From the graph it can be seen that MR = MC = 55 when q = 8000.
To check if this gives maximum profit
For q = 7000, MR = 56 and MC = 54, since MR > MC adding more q would increase the profit.
For q = 9000, MR = 54 and MC = 58, since MR < MC, the profit is negative for adding additional products.
Thus q = 8000 maximizes the profit
Text Problem 4.5.15
Given that C’(500) = 75 and R’(500) = 100.
Since extra cost incurred in increasing q from 500 is less than the extra revenue earned from increasing q from 500, this would indicate that production increased from q = 500 would yield greater profit.
Explaining,
The extra cost incurred in increasing q to 501 would be $75 and the extra revenue earned would be $100. Thus the additional profit for that 1 item would be = $100 - $75 = $25.
Thus the quantity produced should be increased to increase the profits.
Text Problem 4.5.18
Given cost function C(q) = 0.01q^3 - 0.6q^2 + 13q
(a)
Fixed cost is the value of the cost function when q = 0
C(0) = 0 + 0 + 0 = 0
Thus the function indicates that there is no fixed cost.
(b)
Suppose q quantities is produced and sold.
The revenue earned from selling q quantities will be R(q) = 7q
The cost for the same q quantities is C(q) = 0.01q^3 - 0.6q^2 + 13q
Thus the profit earned pi(q) = 7q - 0.01q^3 + 0.6q^2 - 13q = -0.01q^3 +0.6q^2 - 6q
For maximum profit pi’(q) should be 0
Using the rules of differentiation we get
pi’(q) = -0.03q^2 + 1.2q - 6
and p”(q) = -0.06q + 1.2
for critical points
pi’(q) = 0
0.03q^2 - 1.2q + 6 = 0
q^2 - 40q + 200 = 0
On solving the quadratic we get
q = 20 + 10sqrt(2) and 20 - 10sqrt(2)
The value of the second derivative at q = 20 + 10sqrt(2) is negative, thus q = 20 + 10sqrt(2) indicates a point of maxima and q = 20 - 10sqrt(2) indicates a point of local minima.
Thus an approximate value of q = 34 gives the maximum profit
Revenue for 34 goods = 34 * $7 = $238
The cost for the same is C(34) = 393.04 - 693.6 + 442 = $141.44
Thus the profit for the same is $238 - $141.44 = $96.56
(c)
Let x be the increase in price.
Thus the selling price of the goods will be $7 + x
The quantity of goods sold will be 34 - 2x
The profit should be
pi(x) = (34 - 2x)*(7 + x) - C(34)
pi(x) = (34 - 2x)*(7 + x) - $141.44
To maximize profit pi’(x) = 0
pi’(x) = (34 - 2x) + (7 + x)*(-2) = 0
34 - 2x - 14 - 2x = 0
4x = 20
Thus x = 5
Thus an increase of $5 in the selling price should be done.
Text Problem 4.5.21
Given that R(x) is the return function and r(x) is the return per dollar invested.
Thus r(x) = R(x) / x
(a)
r(x) function can be rewritten as
r(x) = (R(x) - 0) / (x - 0)
Since the R(x) function starts from the origin
The value of r(x) determines the slope of the line joining the origin to a point (x,R(x)) on the curve of R(x) vs. x.
From the graph it can be clearly seen that this slope will be greatest only when the line joining the origin to the point on the R(x) curve will be tangent to the curve. Thus the value of R(x) which maximizes the value r(x) should be such that the line joining the origin to the point (x,R(x)) will be the tangent to the R(x) curve.
(b)
Since the R(x) function is concave up for certain value of x and then concave up down for further values of x, the graph of r(x) will be increasing first and then decreasing. r(x) function increases at an decreasing rate and then decreases. When plotted we notice that the point at which the r(x) function starts decreasing is the point at which the tangent line from origin meets the R(x) graph.
Thus the r(x) becomes maximum at a point where the slope of the r(x) curve becomes 0.
(c)
Given function r(x) = R(x) / x
Using the rules of differentiation we get
r’(x) = (xR’(x) - R(x))/x^2
For maximum value r’(x) = 0
Since x^2 cannot be positive
xR’(x) = R(x)
Thus R’(x) = R(x)/x = (R(x) - 0)/(x - 0)
This indicates that the line joining origin to the point on the curve of R(x) should be tangent to the R(x) curve.
Text Problem 4.5.24
(a)
The function f(x) determines the number units of item a manufacturing company can produce from x units of raw material.
It is pretty obvious that more raw material would lead to more units of item manufactured. Thus x increases, f(x) should monotonically increase. Since the f(x) function is increasing for all values of x, f’(x) must be positive for all value of x.
(b)
Given that $p is the amount at which 1 unit is sold.
Thus if f(x) units are sold the total amount earned by = $pf(x)
Given that $w is the amount of 1 unit of raw material
If the company uses x units of raw material, the total cost of production would be = $wx
Thus the profit = revenue - cost
pi(x) = pf(x) - wx
(c)
f’(x*) indicates the number of units manufactured from the amount of the raw material that maximizes the profit.
For maximum profit MR = MC
Thus pf’(x) = w
f’(x) = w/p
Thus x* = f’^-1(w/p)
And thus f’(x*) = w/p
(d)
The revenue function R(x) for the company = pf(x)
Near x*, the revenue function will have slope decreasing as x increases. Thus the second derivative of the revenue function will be positive. Since p is positive, f”(x) should be negative near x*.
Thus f”(x*) will be negative.
(e)
We know that
f’(x*) = w/p
Differentiating both sides with respect to w we get
f”(x*) * d(x*)/dw = 1/p
Thus the formula for
d(x*)/dw = 1 / (pf”(x*))
From (d) we know that f”(x*) is negative and p is positive.
Thus d(x*)/dw should be negative.
(f)
From the answer in part (e) we can see that d(x*)/dw is negative, that is the value of x* decreases with increase in w.
If the price of w, that is the value of w goes up, the value of x* will come down. This indicates that, in order to maximize profit the company need to buy lesser raw materials as compared to what it used to buy.
Text Problem 4.5.26
(a)
The graph of the straight line x + 2y = 10 is a straight line passing through (0, 5) and (10,0).
The graph of the function 10xy = 100 will be a hyperbolic curve in the first quadrant cutting the straight line x + 2y = 10 at 2 different points.
Graph of 10xy = 200 will again be a hyperbolic curve such that it moves away from the curve 10xy = 100. Thus curve 10xy = 200 does not touch or cut the line x + 2y = 10. That is no value of x and y satisfy both the equations simultaneously.
The graph of the function 10xy = 300 will again be a hyperbolic further moving away from the previous curve and the straight line, not touching or intersecting the straight line x + 2y = 10.
(b)
Given that the quantity function is Q = 10xy
And the cost function is C = x + 2y thousand dollars
Thus x = C - 2y
Substituting the x value in the quantity function we get
Q = 10(C - 2y)y
Now to maximize the production we need to maximize the value of Q.
Since the value of the Q function will increase as the value of C increases, the maximum value of Q will occur when C is 10 since that is the maximum value that C be attain.
Thus the maximum production occurs at a point where the production curve is tangent to the line C = 10, that is x + 2y = 10
(c)
Given that Q = 10(C - 2y)y
Since for maximum value of Q we have C = 10 thus
Q = 10(10 - 2y)y
Using the rules of differentiation we get
Q’ = 10(10 - 2y) + 10y(-2)
From maximum value of Q, Q’ = 0
Thus we have
10 - 2y - 2y = 0
10 = 4y
y = 5/2 units
Since we have 10 = x + 2y and y = 5/2 units
We get
10 = x + 5
And x = 5 units
Thus the value of Q will be Q = 10(5)(5/2) = 125 sq units.
Thus the maximum production is 125 sq units
Text Problem 4.5.32
The would generally want to
(a) Maximize revenue
(b) Maximize marginal revenue
(c) Minimize cost
(d) Minimize marginal cost.
All the above factors will be favorable to the company in maximizing its production.
Text Problem 4.5.35 and Text Problem 4.5.37 not present in the textbook.
Text Section 4.6
Text Problem 4.6.1
Given temperature function
H = 4 + 16e^(-0.02t) C
The rate at which the temperature of the body is changing will be equal to the derivative of this function
H’ = 0 + 16e^(-0.02t) * (-0.02) C/min
The rate at which the temperature is changing initially will be equal to H’(0) which is
H’(0) = 0 + 16e^(0) * -0.02 = -0.32 C/min
After 10 min the value of the derivative is
H’(10) = 0 + 16e^(-0.02*10)*(-0.02) C/min
H’(10) = -0.32e^(-0.2) = approximately -0.262 C/min
Text Problem 4.6.3
Given power function
P = 81/R 1/ohm
Using the rules of differentiation the derivative of the function is
P’ = (81/R)’ = -81/(R^2) 1/(ohm^2)
The rate of power change with respect to resistance will be
P’ = -81/(R^2) 1/(ohm)^2
Text Problem 4.6.6
Given pressure function
P = 30(e^(-3.23 * 10^-5h))
(a)
When P = 25 inches
25 inches = 30(e^(-3.23 * 10^-5*h))
0.834 = e^(-3.23 * 10^-5*h)
ln(0.834) = -0.181521 = -3.23 *10^-5 * h
h = 0.181521 / (3.23 * 10^-5) = 0.056198 * 10^5 feet
Thus at a height of 5.6198 * 10^3 feet above the sea level pressure is 25 inches.
(b)
Given that dP/dt = 0.1 inches/min
Using the rules of differentiation, the derivative of the function is
P’ = 30(e^(-3.23 * 10^-5 * h) * (-3.23 * 10^-5*(h’))
0.1 = 25 * (-3.23 * 10^-5 * (h’))
0.1 = -80.75 * 10^-5 * h’
h’ = -0.0012384 * 10^5 feet/sec
Thus the rate at which the altitude is changing is -1.2384 * 10^2 feet/sec
Text Problem 4.6.9
Given area function A = (9/16)*(4theta - sin(4theta))
Given d(theta)/dt = 0.2 radians/min when theta = pi/4
Using the rule of differentiation we get
dA/dt = (9/16)*(4theta - sin(4theta))’ = (9/16)*(4(d(theta)/dt) - cos(4theta)*(4(d(theta)/dt))
The value of this derivative at theta = pi/4 and d(theta)/dt = 0.2 radians/min is
dA/dt = (9/16)*(4*0.2 - cos(pi)*(4*0.2))
dA/dt = (9/16)*(0.8 + 0.8) = 0.9 cm^2 / min
Thus when theta = pi/4 and the d(theta)/dt = 0.2 radians/min the rate at which the area is increasing is 0.9 cm^2/min
Text Problem 4.6.12
Given x^2 + y^2 = 25
dx/dt = 6
(a) When x = 0
y^2 = 25
Differentiating the given equation we get
2x(dx/dt) + 2y(dy/dt) = 0
0 + 2y(dy/dt) = 0
Thus dy/dt = 0
(b) When x = 3
9 + y^2 = 25
y^2 = 16
y = 4, y is to kept positive
Differentiating the given equation we get
2xdx/dt + 2ydy/dt = 0
3(6) = - ydy/dt
y(dy/dt) = -18
When y = 4
dy/dt = -18/4 = -4.5
(c)
x^2 + y^2 = 25
x = 4
16 + y^2 = 25
y^2 = 9
y = 3, y is to be kept positive.
Differentiating the given equation we get
2x(dx/dt) + 2y(dy/dt) = 0
4*(6) = -3dy/dt
dy/dt = -18/3 = -6
Thus dy/dt when y is positive is -6 when x = 4 and dx/dt = 6
Text Problem 4.6.16
Given that the length of one = a = 10 cm
The length of the other side = b = 12 cm
The rate at which the other side b is increasing = db/dt = 3 cm/min
The area of the rectangle will thus be A = area = ab
Differentiating the given area equation we get
dA/dt = a*(db/dt) = 10 cm * 3 cm/min = 30 cm^2/min
Thus the area of the rectangle is increasing at the rate of 30 cm^2/min when one side of the rectangle has a length of 10 cm and the other side has a length of 12 cm and is increasing at the rate of 3 cm/min
Text Problem 4.6.17
Given the length of one side of the rectangle a = 8 cm
The length of the other side be b
Let the length of the diagonal be d
Using Pythagoras theorem we get d^2 = a^2 + b^2 = 64 + b^2
Given that db/dt = 3 cm/min when b = 6 cm
When a = 8 and b = 6, d will be equal to
d^2 = 64 + 36 = 100, d = 10 (negative solution neglected since length cannot be negative)
Differentiating the diagonal function we get
2d(d’) = 0 + 2b(b’)
d’ = b*(b’/d) = 6*3/10 = 1.8 cm/unit
Thus the diagonal is changing at the rate of 1.8 cm/unit
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