Solve for X Need help ASAP
1 answer:
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This problem is a problem using
d = r * t
First part of the trip
d = 96 miles
r = ??
t = ??
Second part of the trip
d = 15 miles
r = r - 5
t = 4 - t
Solve for r first.
96/r + 15/(r - 5) = 4 hours multiply through by r and r - 5
96(r - 5) + 15r = 4*(r - 5)(r) Remove the brackets on the left
96r - 480 + 15r = 4*r*(r - 5) Collect like terms on the left.
111r - 480 = 4r(r - 5) Remove the brackets on the right.
111r - 480 = 4r^2 - 20r Bring the left side over to the right side.
0 = 4r^2 - 20r - 111r + 480 Collect like terms.
0 = 4r^2 - 131r + 480
The only way I could do this was to use the quadratic equation.
x = (131 + 97.38)/8
x = 28.55
x = (131 - 97.37)/8
x = 4.20
4.2 isn't going to work because the slower speed has to be larger than 5, otherwise you will get a negative speed (as this solution will give), so 4.20 is an extraneous solution.
r = 28.55 for the first part of the trip and
r = 23.55 for the second part of the trip.
Check.
Let's see if the times add up to 4
d/r = t
96/28.55 = 3.36
15/22.75 = 0.64
The times do = 4, so the rates are correct.
Answers
r1 = 28.55 for the 96 mile trip.
r2 = 23.55 for the 15 mile trip.
d = r * t
First part of the trip
d = 96 miles
r = ??
t = ??
Second part of the trip
d = 15 miles
r = r - 5
t = 4 - t
Solve for r first.
96/r + 15/(r - 5) = 4 hours multiply through by r and r - 5
96(r - 5) + 15r = 4*(r - 5)(r) Remove the brackets on the left
96r - 480 + 15r = 4*r*(r - 5) Collect like terms on the left.
111r - 480 = 4r(r - 5) Remove the brackets on the right.
111r - 480 = 4r^2 - 20r Bring the left side over to the right side.
0 = 4r^2 - 20r - 111r + 480 Collect like terms.
0 = 4r^2 - 131r + 480
The only way I could do this was to use the quadratic equation.
a = 4
b = - 131
c = 480
x = (131 + 97.38)/8
x = 28.55
x = (131 - 97.37)/8
x = 4.20
4.2 isn't going to work because the slower speed has to be larger than 5, otherwise you will get a negative speed (as this solution will give), so 4.20 is an extraneous solution.
r = 28.55 for the first part of the trip and
r = 23.55 for the second part of the trip.
Check.
Let's see if the times add up to 4
d/r = t
96/28.55 = 3.36
15/22.75 = 0.64
The times do = 4, so the rates are correct.
Answers
r1 = 28.55 for the 96 mile trip.
r2 = 23.55 for the 15 mile trip.
Answer:
The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.
Step-by-step explanation:
Volume of the Cylinder=400 cm³
Volume of a Cylinder=πr²h
Therefore: πr²h=400
Total Surface Area of a Cylinder=2πr²+2πrh
Cost of the materials for the Top and Bottom=0.06 cents per square centimeter
Cost of the materials for the sides=0.03 cents per square centimeter
Cost of the Cylinder=0.06(2πr²)+0.03(2πrh)
C=0.12πr²+0.06πrh
Recall:
Therefore:
The minimum cost occurs when the derivative of the Cost =0.
r=3.17 cm
Recall that:
h=12.67cm
The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.