Math 103. Homework 7. Solutions

1.[61, p. 221] Find an equation of the tangent line to the graph of the function

f(x) = x   _____ Ö2x2+7

at the point (3,15).

Solution. The slope of such line is given by the derivative f¢(3). We compute f¢(x) using the product rule and chain rule:

f¢(x) = (2x2+7)1/2 + x(1/2)(2x2+7)-1/2 (4x) = (2x2+7)1/2 + 2x2(2x2+7)-1/2

Plugging in x=3into the derivative:

f¢(3) = (2(3)2+7)1/2 + 2(3)2(2(3)2+7)-1/2 = 43 5

An equation for the line passing through (3,15) with slope 43/5 is

y = 43 5 x - 54 5

2. [13, p. 237] The Pulsar corporation manufactures certain model of 19-inch color TV sets. The quantity x of these sets demanded each week is related to the wholesale unit price p by the equation

p=-0.006 x +180

The weekly total cost incurred by Pulsar for producing x sets is

C(x) = 0.000002 x3 -0.02 x2 +120 x+60,000

dollars.

(a) Find the revenue function R and the profit function P.
(b) Find the marginal cost function C¢, the marginal revenue function R¢ and the marginal profit function P¢.
(c) Compute C¢(1000), R¢(1000) and P¢(1000) and interpret your results.

R(x)=xp(x) = x(-0.006 x +180) = -0.006 x2 +180 x

and the profit function P is

P(x) = R(x)-C(x) = -0.006 x2 +180 x-( 0.000002 x3 -0.02 x2 +120 x+60,000 ) = -0.000002 x3 + 0.014 x2+60 x -60,000

(b) C¢(x) = 0.000006 x² -0.04 x +120

R¢(x)=-0.012 x+180

P¢(x) = -0.000006 x² +0.028 x +60

(c) C¢(2000) = 0.000006 (2000)² -0.04 (2000) +120 = 64, and this says that at the level of production of 2000 units, the cost of producing the 2001st unit is $64

R¢(2000)=-0.012 (2000)+180 = 156, and this says that the revenue realized in selling the 2001st unit is $156

P¢(2000) = -0.000006 (2000)² +0.028 (2000) +60=92, and this says that the profit realized in selling the 2001st unit is $92

3. [15, p. 237] Find the average cost function

C

associated with the total cost function of the previous problem.

(a) Find the marginal average cost function




C   ¢


(b) Compute




C   ¢(5000)       and        C   ¢(10,000)

and interpret your results.

Solution. (a) The average cost function is

C   (x) = C(x) x = 0.000002 x3 -0.02 x2 +120 x+60,000 x = 0.000002 x2 -0.02 x +120 + 60,000 x

and the marginal average cost function is

C   ¢(x) = 0.000004 x -0.02 - 60000 x2

(b)

C   ¢(5000) = 0.000004(5000) -0.02 - 60000 (5000)2 = -0.0024

and this says that, at the level of production of 5000 units, the average cost of production is dropping 2/10 penny/unit.

C   ¢(10000) = .000004(10000)-0.02 - 60000 (10000)2 = 0.0194

which says that the average cost is increasing when 10,000 units are produced.

4. The management of the Titan Tire Company has determined that the quantity demanded x of their Super Titan tires per week is related to the unit price by the equation

p = 144-x2

where p is measured in dollars and x in units of a thousand.

(a) Compute the elasticity of the demand when p= 63 and 108.
(b) Interpret the results obtained in part (a).
(c) For what values of p is the demand unitary?

Solution. First write the demand equation as a function of p, that is, solve for x and get

f(p) = (144-p)1/2

The derivative f¢(p) is computed using the chain rule

f¢(p) = (1/2)(144-p)-1/2(-1)

The elasticity of demand is given by

E(p) = - pf¢(p) f(p) = - p(-1/2)(144-p)-1/2 (144-p)1/2 = p 2(144-p)

(a) Plug in p=63 and p=108 into E(p) and get

E(63) = 63 2(144-63) » 0.39

E(108)= 108 2(144-108) = 1.5

(b) When the price is set at $63, an increase of $1 in price per tire will result in a decrease of approximately 0.39 percent in the quantity demanded. When the price is set at $108, an increase of $1 in price per tire will result in a decrease of approximately 1.5 percent in the quantity demanded.

(c) The demand is unitary when E(p)=1. Thus we have to solve the equation

p 2(144-p) =1

or

p = 2(144-p) = 288 -2p

or

p = 288 3 = 96

5. [30, p. 238] The quantity demanded per week x (in units of a hundred) of the Mikado miniature camera is related to the unit price p (in dollars) by the demand equation

5p = 400-x2        (0 £ p £ 80)

(a) Is the demand elastic or inelastic when p=40? When p=60?
(b) When is the demand unitary?
(c) If the unit price is lowered slightly from $60, will the revenue increase or decrease?
(d) If the unit price is increased slightly from $40, will the revenue increase or decrease?

Solution. First write the demand equation as x=f(p), quantity demanded x as function of unit price p:

x=f(p) =   ______ Ö400-5p

Then

f¢(p) = -5 2   ______ Ö400-5p

and

E(p)=-p f¢(p) f(p) = -p -5 2   ______ Ö400-5p     ______ Ö400-5p = 5p 800-10p

(a)

E(p) = 5(40) 800-10(40) = 1 2

so the demand is inelastic when p=40.

E(60) = 5(60) 800-10(60) = 1.5

so the demand is elastic when p=60.

(b) The demand is unitary if

E(p) = 5p 800-10p =1

or 5p=800-10p. That is, the demand is unitary when p=160/3.

For parts (c) and (d), note the following relation. The revenue function R(p) as function of unit price is:

R(p) = p f(p)

Thus the marginal revenue is

R¢(p) = f(p) + p f¢(p) = f(p) [1+p f¢(p) f(p) ] = f(p)[1-E(p)]

So if the demand is elastic at p (E(p) > 1), then an increase in unit price will cause the revenue to decrease, whereas a decrease in the unit price will cause the revenue to increase.

Analogously, if the demand is inelastic at p (E(p) < 1), then R¢(p) > 0, so that revenue is decreasing at p. A slight increase in the unit price will make the revenue to increase, while a little decrease in unit price will cause the revenue to decrease.

(c) Since the demand is elastic at p=60, lowering the unit price a little will cause the revenue to increase.

(d) Since the demand is inelastic at p=40, a slight increase of the unit price will cause the revenue to increase.