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

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

answer: ur mum is gay

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1