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german
3 years ago
9

A car rental agency rents 440 cars per day at a rate of ​$30 per day. For each ​$1 increase in​ rate, 10 fewer cars are rented.

At what rate should the cars be rented to produce the maximum​ income? What is the maximum​ income
Mathematics
2 answers:
tigry1 [53]3 years ago
8 0

Answer:

Maximum cost is 37, and the maximum income is 370*70 = 25.900

Step-by-step explanation:

Let y be the number of cars that are rented, and let b  the rate at which it is charged the rental per day. Then, the income of the car rental is y*b. We know that if b= 30 then y = 440. Let k be the increase in rate. So, we know that if k=1, then y decreases in 10 cars (that is y = 440-10). Let b = 30+k be the new renting rate. In this case, we have that y = 440-10k, since for each dolar we increment the renting rate, the number of rented cars reduces in 10. Then, the income as a function of k is given by

I(k) = (440-10k)\cdot(30+k)

We want to find k such that I is maximum. Then, we will find it's derivative and solve it equal to 0. The derivative is given by

I'(k) = (-10)\cdot (30+k) + (440-10k) = 440-300-20k = 140-20k = 20(7-k)

Those, the equation I'(k) =0 has the solution k=7.

Note that I''(k) = -20<0, so this implies that k=7 is a maximum by the second derivative criteria.

Hence, the optimum rate is 30+7 = 37 and the number of cars that will be rented is 440-70 = 370. Then, the income is I(7) = 25900

Ratling [72]3 years ago
5 0

Answer:

Maximum cost C is

C = P(740-10P) = 37(740-370) = 37 × 370 = $13,690

C = $13,690

Step-by-step explanation:

Let Q represent the number of cars rented per day and P the rate.

To form the demand equation;

Q = a - bP

A car rental agency rents 440 cars per day at a rate of ​$30 per day

440 = a - 30b ......1

For each ​$1 increase in​ rate, 10 fewer cars are rented

(440-10) = a - (30+1)b

430 = a - 31b ......2

Subtracting equation 2 from 1

10 = b

From equation 1;

440 = a - 10(30)

a = 440 + 300 = 740

So;

Q = 740 - 10P

Cost C = Q×P = P (740-10P) = 740P - 10P^2

Maximum cost is at dC/dP = 0

dC/dP = 740 - 20P = 0

P = 740/20 = $37

Maximum cost C is

C = P(740-10P) = 37(740-370) = 37 × 370 = $13,690

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