Respuesta :
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
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
[tex]I(k) = (440-10k)\cdot(30+k)[/tex]
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
[tex]I'(k) = (-10)\cdot (30+k) + (440-10k) = 440-300-20k = 140-20k = 20(7-k)[/tex]
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
