Stuck on a math question. Please help!?

Question

Here is the problem:
While traveling x miles per hour, a truck burns fuel at the rate of G(x) gallons per mile, where
G(x) = 1/32 (64/x + x/50)

a) If fuel costs $2.88 per gallon, find the speed that will produce the minimum total cost for a 400-mile trip.
b) Find the minimum total cost.

Any help is appreciated! TIA.

Answer 1

well,

you have to find the minimum of your G(x) function.

for that, G'(x) = 0

G'(x) = 1/32 (-64 / x^2 + 1/50) =
-2 / x^2 + 1/1600

G'(x) = 0 ==>
-2 / x^2 + 1/1600 = 0 ==>
2/x^2 = 1/1600 ==>
2 = x^2 / 1600 ==>
x^2 = 3200 ==>
x = 56.57 mph

G(x) =
1/32 (64/x + x/50) =
1/32 (64/ 56.57 + 56.57 /50) =
0.0707 gallons per mile

since we're travelling 400 miles, at $2.88 a gallon, the answer is

0.0707 * 400 * 2.88 = $81.46

Answer 2

Well, first I'd multiply that 1/32 through to simplify things

G(x) = 2/x + x/1600

Now take the derivative
-2/x^2 + 1/1600

Set the derivative equal to zero and solve for x
-2/x^2 + 1/1600 = 0
-2/x^2 = -1/1600
3200 = x^2
x ~= 56.57

Cost at 56.57 mph:
2/56.57 + 56.57/1600 = 0.0707 gallons per mile
times 400 miles = 28.28 gallons
times $2.88 per gallon = $81.46

Test 1: Try 55 mph
2/55 + 55/1600 = 0.0738 gallons per mile --> more expensive

Test 2: Try 60 mph
2/60 + 60/1600 = 0.0783 gallons per mile --> more expensive

The tests show that we have indeed located the minimum

Answer 3

a)
We have to find the value of x (speed) at which G(x) (gallons) is minimum,
for that, find the derivative of g(x), equate it to zero, and solve for x, ie
G'(x) = 1/32(-64/x^2+1/50) = 0

b)
substitute the above x in G(x). This is the optimum gas consumption per mile, let's call it as 'gpm'. Now 400/gpm gives the total gas consumed during the 400 mile trip and 2.88 * 400/gpm gives the minimum cost!!