Question

4. Mateo's room overlooks a CFL football stadium. He decides to rent a telescope for $50.00 a week and charge his friends to use it to peep at the games for 30 seconds. He can act as a single-price monopolist for renting out “peeps.” For each person who takes a 30-second peep, it costs Mateo $0.20 to clean the eyepiece. This table shows the information Mateo has gathered about the weekly demand for the service. peep $1.20 1.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 demanded 0 100 150 200 250 300 350 400 450 500 550 a. For P = $0.70, $0.60, and $0.50, calculate the total revenue from selling peeps and the marginal revenue per peep. P = $0.70,TR = $ P = $0.60,TR = $ P = $0.50, TR = $

Answer 1

a)

Price	Qd	TR = P*Q	MR = ∆TR/∆Q	MC
$1.20	0	$120*0 =$0	---	$0.20
$1.00	100	$1*100 =$100	($100 - $0)/(100-0)=$100/100 =$1	$0.20
$0.90	150	$0.90*150 =$135	($135-$100)/(150-100) =$35/50 = $0.70	$0.20
$0.80	200	$0.80*200=$160	($160-$135)/(200-150) = $25/50 =$0.50	$0.20
$0.70	250	$0.70*250=$175	($175-$160)/(250-200) = $15/50 = $0.30	$0.20
$0.60	300	$0.60*300 = $180	($180 - $175)/(300-250) =$5/50 = $0.10	$0.20
$0.50	350	$0.50*350 = $175	($175 - $180)/(350 -300) = -$5/50 = -$0.10	$0.20
$0.40	400	$0.40*400=$160	($160 - $175)/(400-350) = -$15/50 = -$0.30	$0.20
$0.30	450	$0.30*450 =$135	($135-$160)/(450-400) =-$25/50 = -$0.50	$0.20
$0.20	500	$0.20*500 =$100	($100-$135)/(500-450) =-$35/50 = -$0.70	$0.20
$0.10	550	$0.10*550 =$55	($55-$100)/(550-500)=-$45/50=-$0.90	$0.20

Therefore, when P = $0.70 , TR = P*Q = $0.70*250 = $175

When P = $0.60 , TR =P*Q = $0.60*300 = $180

When P = $0.50, TR =P*Q = $0.50*350 = $175

MR when P = $0.70 to $0.60 is ∆TR/∆Q = ($180-$175)/(300-250)=$5/50 = $0.10

MR when P = $0.60 to $0.50 is ∆TR/∆Q = ($175-$180)/(350-300) = -$5/50 = -$0.10

​​​​​​b) Profit maximising condition of Mateo will be where we will get MR = MC. If MR is not equal to MC we will take the MR which is close to MC but MR must be greater than MC. So, profit maximising level will be there where MR is just above the MC if we do not get exact MR = MC

From the table what we have calculated in part (a) we can get from there there is no MR which is exactly equal to MC. Because MC is fixed at $0.20 per unit. MR we get $0.30 and $0.10 when price is $0.70 and Q =250 and P =$0.60 and Q =$300

So, we will not take MR =$0.10 because it is less than MC as MC = $0.20.

So profit will be maximises when Q = 250. Price charged at this level of quantity is P = $0.70

Profit at this Q =250 and P =$0.70 is TR - TC = ($0.70*250)-(0.20*250)-$50 = $175 - $50 - $50 = $75.  

Profit maximising quantity = 250 (Ans)

Profit maximising price = $0.70(Ans)

Maximal Profit = $75 (Ans).

(c). Due to paying extra $0.20 to landlady for every peep he sells then as a result of this Mateo's marginal cost will rise by $0.20.

Ans. Option -1 : Mateo's marginal cost per peep will increase by $0.20.

New profit maximising quantity will be based on new marginal cost of Mateo. Due to paying $0.20 per peep to the landlady the new marginal cost becomes $0.20 + $0.20 = $0.40.

As MC = $0.40 we will again look into the MR which is close to MC =$0.40 and MR = MC or MR is just above the MC.

From the table in part (a) we can get MR = $0.50 at Q = 200 and MR = $0.30 at Q = 250.

So, we will take Q = 200 because at this level MR = $0.50 is just above the new MC i.e $0.40.

New equilibrium Quantity = 200 and price = $0.80

New Profit maximising quantity Q = 200(Ans)

New Profit maximising price P = $0.80

New Profit = TR - TC = $0.80*200 - $0.40*200 - $50 = $160 - $80 - $50 = $30. (Ans).