Question

A car rental agency has 150 cars. The owner finds that at a price $48 per day he can rent all the cars. For each $2 increase in price, the demand is less and 4 fewer cars are rented. What rental charge will maximize revenue?

Answer 1

Let us make an equation for demand as a function of price.

d(p)=(48,150), (50, 146)

So the rate of change is (146-150)/(50-48)=-4/2=-2. So the slope of this function is -2 and the initial value. Now we have:

d(p)=-2p+b and we know we have the point (50,146)

146=-2(50)+b

146=-100+b

246=b so

d(p)=-2p+246

Revenue is price times demand so:

r(p)=p*d(p)

r(p)=p(-2p+246)

r(p)=-2p^2+246p

Revenue is maximized when dr/dp=0

dr/dp=-4p+246

dr/dp=0 when 4p=246

p=$61.50

So a rental charge of $61.50 maximizes revenue.

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In case you were curious...

revenue max=r(61.5)=$7564.50

cars rented at this rad=d(61.5)=123 cars