For speeds between 40 and 65 miles per hour, a truck gets 480/x miles per gallon when driven at a constant speed of x miles per hour. Diesel gasoline costs $2.23 per gallon, and the driver is paid$15.10 per hour. What is the most economical constant speed between 40 and 65 miles per hour at which to drive the truck?

I would just like to know if I calculated this right...thanks!

D(S)=P/S
C(S) ?= G/(480/S)
C(48) ?= G/(480/48)
C(48) ?= G/10
At 10 MPG, each mile costs 1/10 of a gallon of gas
C(S)=SG/480
F(S)=P/S+SG/480
F'(S)= -P/S^2+G/480
P/S = PS^(-1), DS/DY(PS^(-1))=-1PS^(-2)
DS/DY(SG/480)=G/480
Min[F'(S)] @ 0=-P/S^2+G/480, solved for S

P/S^2 = G/480
S^2/P = 480/G
S^2 = 480P/G

S=|SQRT(480P/G)
For P=15.10 and G=2.23, we get S=SQRT(480*15.1/2.23)= ~57.011 MPG.

2. Originally Posted by lisa1984wilson
For speeds between 40 and 65 miles per hour, a truck gets 480/x miles per gallon when driven at a constant speed of x miles per hour. Diesel gasoline costs $2.23 per gallon, and the driver is paid$15.10 per hour. What is the most economical constant speed between 40 and 65 miles per hour at which to drive the truck?

I would just like to know if I calculated this right...thanks!

D(S)=P/S
C(S) ?= G/(480/S)
C(48) ?= G/(480/48)
C(48) ?= G/10
At 10 MPG, each mile costs 1/10 of a gallon of gas
C(S)=SG/480
F(S)=P/S+SG/480
F'(S)= -P/S^2+G/480
P/S = PS^(-1), DS/DY(PS^(-1))=-1PS^(-2)
DS/DY(SG/480)=G/480
Min[F'(S)] @ 0=-P/S^2+G/480, solved for S

P/S^2 = G/480
S^2/P = 480/G
S^2 = 480P/G

S=|SQRT(480P/G)
For P=15.10 and G=2.23, we get S=SQRT(480*15.1/2.23)= ~57.011 MPG.
I get 57.01 mph

Let D be the distance traveled in miles and x the speed in miles per hour. The time of travel is
t = D/x and the driver is paid $15.10D / x. The number of gallons of gasoline is D miles / (480/x miles/gallon) = Dx / 480 gallons. The cost of travel is C = 15.1D / x 2.23Dx/480 4. Originally Posted by lisa1984wilson The teacher said this was wrong and this is what she said....please help! is still way off....Start with Let D be the distance traveled in miles and x the speed in miles per hour. The time of travel is t = D/x and the driver is paid$15.10D / x. The number of gallons of gasoline is
D miles / (480/x miles/gallon) = Dx / 480 gallons.
The cost of travel is C = 15.1D / x 2.23Dx/480
Which is wrong the cost of travel is:

$\displaystyle C(x,D)=\frac{15.10 D}{x} + \frac{2.23 D x}{480}$

You can set $\displaystyle D=1$ as everything on the right is proportional to $\displaystyle D$ and the optimum speed is the same whatever the distance.

Then you need to solve:

$\displaystyle \frac{d}{dx} C(x,1)=-\frac{15.10}{x^2}+\frac{2.23}{480}=0$

which appears to give $\displaystyle x=57.0107$ mph

CB

5. Originally Posted by lisa1984wilson
For speeds between 40 and 65 miles per hour, a truck gets 480/x miles per gallon when driven at a constant speed of x miles per hour. Diesel gasoline costs $2.23 per gallon, and the driver is paid$15.10 per hour. What is the most economical constant speed between 40 and 65 miles per hour at which to drive the truck?

I would just like to know if I calculated this right...thanks!

D(S)=P/S
C(S) ?= G/(480/S)
C(48) ?= G/(480/48)
C(48) ?= G/10
At 10 MPG, each mile costs 1/10 of a gallon of gas
C(S)=SG/480
F(S)=P/S+SG/480
F'(S)= -P/S^2+G/480
P/S = PS^(-1), DS/DY(PS^(-1))=-1PS^(-2)
DS/DY(SG/480)=G/480
Min[F'(S)] @ 0=-P/S^2+G/480, solved for S

P/S^2 = G/480
S^2/P = 480/G
S^2 = 480P/G

S=|SQRT(480P/G)
For P=15.10 and G=2.23, we get S=SQRT(480*15.1/2.23)= ~57.011 MPG.
You won't get full marks for this becaise:

1. You don't explain what each variable/symbol means before or when you use it.

2. You don't explain what you think you are doing at each major step.

What you are presenting is just a jumble of calculation and leaving it to the reader to sort out what they think you are trying to do.

CB

6. look at the problem in terms of units ...

cost for the truck ...

$\displaystyle \frac{2.23 \, dollars/gal}{(480/x) \, miles/gal} = \frac{2.23x}{480} \, dollars/mile$

cost for the driver ...

$\displaystyle \frac{15.10 \, dollars/hr}{x \, miles/hr} = \frac{15.10}{x} \, dollars/mile$

total cost as a function of speed in units of dollars per mile ...

$\displaystyle C(x) = \frac{2.23x}{480} + \frac{15.10}{x}$

$\displaystyle C'(x) = \frac{2.23}{480} - \frac{15.10}{x^2}$

set $\displaystyle C'(x) = 0$ ...

$\displaystyle \frac{2.23}{480} = \frac{15.10}{x^2}$

$\displaystyle x = \sqrt{\frac{480 \cdot 15.10}{2.23}} = 57.01$ mph

since $\displaystyle C''(x) > 0$ , this value yields a minimum