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A survey shows that 50,000 people will buy a souvenir item if it costs Php 80.00, and that for every Php 2.50 added to the price, 1,000 fewer people will buy the souvenir item. What price (in Php) will give the greatest revenue? Round off your answer to nearest Php.


The answer is 103.

Mc Jason Laurete by Mc Jason Laurete , 23, Philippines

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1 solution

Pariah Dimasupil
Mar 3, 2014

Our constructed formula for the maximum profit is:

π m a x = Δ P ∗ Δ Q \pi_{max}=\Delta P*\Delta Q

where Δ P \Delta P is the change in price being 2.50, and Δ Q \Delta Q is the change in quantity of souvenirs assuming that each buyer only purchases one item. Hence,

π m a x = 2.50 ∗ 1000 = 2500 \pi_{max}=2.50*1000=\boxed {2500}

From this, we want to know the price when the future profit π t + 1 \pi_{t+1} equals this π m a x \pi_{max} .

First, we denote the revenue after the first price increase as T R 1 TR_1 , and it is 82.50 ∗ 49000 = 4 , 042 , 500 82.50*49000=4,042,500 . Profit is:

π 1 = T R 1 − T R 0 \pi_1=TR_1-TR_0

π 1 = 4 , 042 , 500 − 4 , 000 , 000 = 42 , 500 \pi_1=4,042,500-4,000,000=42,500

Consequently, T R 2 TR_2 would be 4 , 080 , 000 4,080,000 and so π 2 = 37 , 500 \pi_2=37,500 . Change in profit is therefore:

Δ π = 37 , 500 − 42500 = − 5000 \Delta \pi=37,500-42500=-5000 .

To determine the future profit π t + 1 \pi_{t+1} , the formula is π t + 1 = π t + Δ π \boxed {\pi_{t+1}=\pi_t+\Delta \pi} . Solving for this, one would arrive at π 9 = 2500 \pi_9=2500 . Meaning that maximum profit and so total revenue is achieved at the 9th price increase. That time, price would be:

P 9 = 80 + ( 2.50 ∗ 9 ) = 80 + 22.5 = 102.5 ≈ 103 P_9=80+(2.50*9)=80+22.5=102.5≈\boxed {103} .

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Easier Solution

Since for every 2.50 2.50 increase in price, the demand decreases by 1000 1000 ; for every unit increase in price (P), the reduction in demand (Q) will be :

1000 / 2.5 = 400 1000/2.5=\boxed{400}

Working backwards from 80 and 50,000 at P=0, the demand will be 400 ∗ 80 + 50000 = 82000 400 * 80 + 50000 = \boxed{82000} .

Now the demand function can be derived as:

Q = 82000 − 400 P Q=82000-400P

Hence, the revenue function will be given as:

P Q = 82000 P − 400 P 2 PQ=82000P-400P^2

Since it is a negative quadratic function, the revenue will be maximized at d P Q / d P = 0 dPQ/dP=0

→ d P Q / d P = d ( 82000 P − P 2 ) / d P = 0 \rightarrow dPQ/dP=d(82000P-P^2)/dP=0

→ 82000 − 800 P = 0 \rightarrow 82000-800P=0

P = 102.5 ≈ 103 P=102.5\approx \boxed{103}

M Akhil - 7 years, 2 months ago

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