Truck Alpha initially costs $413 more than Truck Beta, but it consumes 0.03 liter/km versus 0.06 liter/km for Beta. Both trucks last 9 years, and Beta's salvage value is $144 smaller than Alpha's. Fuel costs $1.71 per liter. Other things being equal, beyond how many kilometers of use per year does Alpha become preferable to Beta?
@DanJS last problem
C(x)_alpha = 413 + 0.03X C(x)_Beta = 269 + 0.06X set equal we get 413 + 0.03X = 269 + 0.06X, X=1600liter/Km 1600liter/km*(1/1.71liter) = 935 km
@DanJS
@DanJS
the question asks when does the cost to operate both trucks have the same value in terms of kilometers, cost as a function of kilometers
C = cost X = kilometer C(X) = some function
Hold on should I first try to determine what the function is as far as what is x and what is y?
k so just have to keep track of these things they tell you , first A has $413 over B A-- C(X) = 413 B-- C(X) =
okay
you want to have the cost in terms of kilometers, that is what they are asking for kilometer value
okay
sorry i am here... the gas for each will cost \[\frac{ X~km }{ 1 }*\frac{ 0.03~L }{ 1~km }*\frac{ $1.71 }{ 1~L}\]
No worries. Let me analyze this for a bit
change the 0.03 to 0.06 for B
cost function, each term should be in dollars, see that has dollars, given the number of km driven X
A-- C(X) = 413 + (1.71*0.03)*X B-- C(X) = (1.71*0.06)*X
Okay i see where you are getting at for a: 0.051x for b: 0.1026x X is in terms of dollars
no X as you defined is km, c(X) is cost in dollars dependent on km driven
Oh I thought the km would cancel but you are referring to x being km so it would be 0.051km*X and 0.1026km*X
B has a smaller sell back value at the end of the 9 years $144, so add that to B A-- C(X) = 413 + (1.71*0.03)*X B-- C(X) = (1.71*0.06)*X + 144
right X is km, notice the 3 terms and how X km cancels from that, leaving $
Yes it leaves dollars. That is why I was assuming the x represented dollars but it doesn't because this is a function of C dollars respect to X a distance
so they are 2 lines, B starts cheaper overall, but increases faster, and will overtake A at some point... (X, C(X))
we tack the 144 on B because it will sell for 144 at the end of it's life? Why wouldn't we take the initital cost and subtract it from the 144 instead to obtain 269 and tack that on instead?
B is $144 less than A value to sell back, B will get 144 less back than A, so add the 144 onto B, or Subtract 144 from A, either is good
okay okay I see so my thoughts were correct but you solved it differently. Okay thanks.
x = 5243.66 km
A-- C(X) = 413 + (1.71*0.03)*X B-- C(X) = (1.71*0.06)*X + 144 i trust ya on that value.. ha
Okay thanks for your help. When are you normally on OS? I am not very active on here usually. I use the site mostly when I get stuck.
actually since it is cost functions, it would be better to put the 144 as -144 on A instead, since they getting money back but since we comparing the relative cost of A and B, either way is fine
usually when i can, sometimes alot, sometimes not for weeks
Okay well I appreciate the explanation to solving the question. I will probably redo the problem on paper and re evaluate what we discussed more.
yeah , that is better than a discussion, i could post the whole solution as one response, but that wouldnt help
I disagree. The discussion helps me understand the problem more. Discussing how to set it up and what do analyze help me the most.
i mean, you writing it down is better, do that ..heh
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