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Nataly [62]
3 years ago
11

On a road trip with a friend, you drive about 70 miles per hour, and your friend drives about 60 miles per hour. The plan is to

drive less than 15 hours and at least 600 miles each day. Your friend will drive more hours than you. How many hours can you and your friend each drive in 1 day?
Mathematics
2 answers:
Kazeer [188]3 years ago
8 0
Hey there,

So. . I hope I'm understanding the question correctly. So . .There's 2 drivers in the car and they basically take turns driving. So when you do 70 by 15 hours, you get about 1,050 and yes, for sure you have past the limit of 600 miles a day. So, this means that they already passes what they had said in the beginning.

Hope this helps.

~Jurgen
Oksanka [162]3 years ago
8 0
Allow x to be equal to your driving time in hours.
<span>allow y to be equal to your friend's driving time in hours. </span>

<span>since rate * time = distance, then your rate times your time equals your distance and your friend's rate times your friend's time equal his or her distance. </span>

<span>your requirements are: </span>

you and your friend's combined driving time has to be less than or equal to 15 hours.
you and your friend's combining driving distance has to be greater than or equal to 600.
<span>the hours that your friend drives must be greater than or equal to the hours that you drive. </span>

<span>an additional constraint not mentioned is that the number of hours that you drive must be greater than or equal to 0. </span>

<span>a graphical solution was chosen as the way to solve this since it is a two dimensional problem with x and y being the only variables. </span>

x is your time.
<span>y is your friend's time. </span>

<span>constraint equations are: </span>

x + y <= 15
70x + 60y >= 600
y >= x
<span>x >= 0 </span>


<span>these equations need to be graphed in order to find the feasible region. </span>

<span>in order to graph these equations, each equation is solved for y. </span>

<span>you get: </span>

y <= 15 - x
y >= (600-70x)/60
y >= x
<span>x >= 0 (my graphing software allows this) </span>


<span>these equations were graphed and the corner points of the feasible region were calculated. </span>

<span>the graph with the feasible region and the corner points is shown below: </span>

 

<span>from this graph, it appears that the limits of the values of x and y are: </span>

0 <= x <= 7.5
<span>4.615 <= y <= 15 </span>

<span>this means that you can drive from 0 to 7.5 hours and your friend can drive from 4.615 to 15 hours. </span>

<span>the corner points on the graph are where the maximum / minimum values of x and y can be found. </span>

<span>anything in between is good as long as the requirements are observed. </span>

<span>for example: </span>

<span>when you drive 0 hours, your friend must drive at least 10 hours so that the minimum distance can be achieved. </span>

<span>when you drive 4.615 hours, your friend must drive 4.615 hours as well for the minimum distance to be achieved. </span>

<span>when you drive 7.5 hours, your friend must drive 7.5 hours as well to satisfy the requirements that your friend must drive at least as many hours as you and your combined time must be less than or equal to 15 hours. </span>

<span>you can pick any hours for you and for your friend as long as the requirements are observed. </span>

<span>for example: </span>

if you pick 5 hours for yourself, then you friend can drive a minimum of 5 hours because your friend's hours must be greater than or equal to yours, or your friend can drive a maximum of 10 hours because you and your friend's time must be less than or equal to 15.
<span>when you drive 5 hours and your friend drives 10 hours, you distance traveled is 70*5 + 60*10 = 350 + 600 = 950 miles which is greater than 600 miles so you've met all the constraints. </span>

<span>you can't drive more than 7.5 hours because your friend must drive at least as much as you and 7.5 + 7.5 = 15, so if you drove more than 7.5, you would both have to be driving more than1 15 which isn't allowed. </span>
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By using the concept of uniform rectilinear motion, the distance surplus of the average race car is equal to 3 / 4 miles. (Right choice: A)

<h3>How many more distance does the average race car travels than the average consumer car?</h3>

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