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choli [55]
3 years ago
8

A car rental agency rents 190 cars per day at a rate of 29 dollars per day. for each 1 dollar increase in the daily rate, 5 fewe

r cars are rented. at what rate should the cars be rented to produce the maximum income, and what is the maximum income?
Mathematics
1 answer:
Leviafan [203]3 years ago
7 0

The equation that we can create from this situation is:

i = (190 – 5 x) * (29 + x)

where i is the income and x is the increase in daily rate

Expanding the equation:

i = 5510 + 190x – 145x - 5x^2

i = -5x^2 + 45x + 5510

Taking the 1st derivative:

di/dx = -10x +45

Set to zero to get the maxima:

-10x + 45 = 0

x = 4.5

 

So the cars should be rented at:

29 + x = 33.5 dollars per day

 

The maximum income is:

i = (190 – 5*4.5) * (33.5)

i = 5,611.25 dollars

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What is the result of subtracting the secound equation from the first -4x-2y=-2 x-2y=9
crimeas [40]

Answer:

When we subtract the second equation  x-2y=9 from the first equation  4x-2y=-2, we get the value of x = 11/5.

Therefore, the required result is:

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Step-by-step explanation:

Given

The first equation

-4x-2y=-2

The second equation

x-2y=9

Let us subtract the second equation from the first equation

-4x-2y=-2

-

\:\underline{x-2y=9}

-4x - 2y - (x - 2y) = -2 - 9

-4x -2y -x + 2y = -2-9

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divide both sides by -5

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Simplify

x=\frac{11}{5}

Thus, when we subtract the second equation  x-2y=9 from the first equation  4x-2y=-2, we get the value of x = 11/5.

Therefore, the required result is:

x = 11/5

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3 years ago
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