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.
2 answers:
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
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
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
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