The minimum of a function in a given domain

[Krushnaraj Pandya]

Homework Statement

The operating cost of a truck is 12+(x/6) per km, when the truck travels x km/hr. If the driver earns 6 rupees (*or whatever currency your country has) per hour, what is the most economic speed to operate the truck on a 400 km road? Also, due to weather, the truck can travel only between 35 and 60 km/hr

Homework Equations

dy/dx =0 and f''(x)<0 for maxima of a function and vice versa.
speed=distance/time

The Attempt at a Solution

12+(x/6) per km into 400 gives total expense and profit per hour is 6 so 6 into 400/x is the profit, so net flow of money is 6(400/x)-400(12+{x/6}). per hour, This'd be maximum when its derivative is 0, taking 400 common and differentiating gives (-6/x^2)-(1/6)=0 which has no solution, I also don't know how to use the condition that 35<x<60. I'd appreciate some help, thank you

 

[DaveE]

Consider another function f(x) = 2x+1 for 2<x<4. What and where is it's maximum value? When is df/dx=0? How does this compare to your problem?

 

[Krushnaraj Pandya]

Consider another function f(x) = 2x+1 for 2<x<4. What and where is it's maximum value? When is df/dx=0? How does this compare to your problem?

Its a strictly increasing function with maximum at around 4, the function in the problem is strictly decreasing so the lower limit will be its maxima and vice versa, I understand perfectly, thank you very much

 

[Ray Vickson]

Homework Statement

The operating cost of a truck is 12+(x/6) per km, when the truck travels x km/hr. If the driver earns 6 rupees (*or whatever currency your country has) per hour, what is the most economic speed to operate the truck on a 400 km road? Also, due to weather, the truck can travel only between 35 and 60 km/hr

Homework Equations

dy/dx =0 and f''(x)<0 for maxima of a function and vice versa.
speed=distance/time

The Attempt at a Solution

12+(x/6) per km into 400 gives total expense and profit per hour is 6 so 6 into 400/x is the profit, so net flow of money is 6(400/x)-400(12+{x/6}). per hour, This'd be maximum when its derivative is 0, taking 400 common and differentiating gives (-6/x^2)-(1/6)=0 which has no solution, I also don't know how to use the condition that 35<x<60. I'd appreciate some help, thank you

First, the constraint does NOT say that $35 < x < 60$; it says that $35 \leq x \leq 60.$ That distinction is very important; it can make the difference between an unsolvable problem and a solvable one.

Next: the "rule" that you should set the derivative to zero when maximizing or minimizing should be viewed with extreme caution when you have restriction on the variables. Often the solution does not satisfy the condition $df/dx = 0$ because the solution occus at an end-point. For example, the solution to the problems $\min/\max x$ subject to $0 \leq x \leq 1$ is at $x=0$ for the min problem and at $x=1$ for the max problem. The derivative equals 1 at both points.

Note, by the way, that the problems $\min / \max x$ subject to $0 < x < 1$ are both unsolvable---they have no solutions at all. That is why it is important to distinguish $\leq$ from $<$ and $\geq$ from $>.$

 

[SammyS]

The Attempt at a Solution

12+(x/6) per km into 400 gives total expense and profit per hour is 6 so 6 into 400/x is the profit, so net flow of money is 6(400/x)-400(12+{x/6}). per hour, This'd be maximum when its derivative is 0, taking 400 common and differentiating gives (-6/x^2)-(1/6)=0 which has no solution, I also don't know how to use the condition that 35<x<60. I'd appreciate some help, thank you

As I understand this problem:
Both the cost of operating the truck, and the cost of paying a driver are expenses for the trucking company. Add those rather than subtracting, and minimize the sum.

 

[Krushnaraj Pandya]

As I understand this problem:
Both the cost of operating the truck, and the cost of paying a driver are expenses for the trucking company. Add those rather than subtracting, and minimize the sum.

In India truck drivers are independent, (stereotypically-punjabi people) so I got the correct answer this way. Thank you for your input though :D .

 

[Krushnaraj Pandya]

First, the constraint does NOT say that $35 < x < 60$; it says that $35 \leq x \leq 60.$ That distinction is very important; it can make the difference between an unsolvable problem and a solvable one.

Next: the "rule" that you should set the derivative to zero when maximizing or minimizing should be viewed with extreme caution when you have restriction on the variables. Often the solution does not satisfy the condition $df/dx = 0$ because the solution occus at an end-point. For example, the solution to the problems $\min/\max x$ subject to $0 \leq x \leq 1$ is at $x=0$ for the min problem and at $x=1$ for the max problem. The derivative equals 1 at both points.

Note, by the way, that the problems $\min / \max x$ subject to $0 < x < 1$ are both unsolvable---they have no solutions at all. That is why it is important to distinguish $\leq$ from $<$ and $\geq$ from $>.$

I understand. Thank you :D

 

[Ray Vickson]

In India truck drivers are independent, (stereotypically-punjabi people) so I got the correct answer this way. Thank you for your input though :D .

When you say "this way", what do you mean? Are you saying that your original formulation in #1 is correct, or are you saying that you agree with that in #5?

What is your final answer, and what is the value you get for maximum profit or minimum cost? I ask, because when I do it I get very bad results.

 

[SammyS]

What is your final answer, and what is the value you get for maximum profit or minimum cost?

I wholeheartedly agree with Ray here.

It's a small, but important, payment to the Homework Helpers to see that the student has understood and has gotten a satisfactory result.

 

[Krushnaraj Pandya]

When you say "this way", what do you mean? Are you saying that your original formulation in #1 is correct, or are you saying that you agree with that in #5?

What is your final answer, and what is the value you get for maximum profit or minimum cost? I ask, because when I do it I get very bad results.

I mean the original was correct.I got 35 km/hr, sorry I didn't mention it.

I wholeheartedly agree with Ray here.

It's a small, but important, payment to the Homework Helpers to see that the student has understood and has gotten a satisfactory result.

I agree as well. An amazing thing to know is, In India the exam-coaching industry is bigger than bollywood, almost every school student goes to a coaching center and the teachers there are paid to teach and remove doubts over such things but they only say "do this, that, use the formula" and dismiss it without much concern for any of the students while the people on PF here care so much more about my learning even though they have nothing to gain from it...Thank you for that :D

 

[Ray Vickson]

I mean the original was correct.I got 35 km/hr, sorry I didn't mention it.

I agree as well. An amazing thing to know is, In India the exam-coaching industry is bigger than bollywood, almost every school student goes to a coaching center and the teachers there are paid to teach and remove doubts over such things but they only say "do this, that, use the formula" and dismiss it without much concern for any of the students while the people on PF here care so much more about my learning even though they have nothing to gain from it...Thank you for that :D

You still have refused to answer two of my questions, even though I had good reasons to ask them. These are
(1) which model are you using---the original "profit" maximization in post #1 or the "cost" minimization proposed in post #5?
(2) what is your resulting minimum cost or maximum profit?

EDIT: I see that you at least hinted at the answer to question (1), although you were not very clear and explicit: you seemed to be saying that you were using model (1), shown below.

When I do it I have either
$$ \max\;\; 6 \left( \frac{400}{x}\right) - 400\left(12+ \frac x 6 \right), \; 35 \leq x \leq 60 \hspace{4ex} (1)$$
or
$$\min\;\; 6 \left( \frac{400}{x} \right) + 400\left(12+ \frac x 6 \right), \; 35 \leq x \leq 60 \hspace{4ex} (2)$$
Problem (1) is the profit maximization problem, while (2) is the cost minimization problem. Both have the solution $x = 35$ (km/hr). The maximum profit is $P_{\max} = - 7064.76$ rupees (a loss of 7064.76 rupees), while the minimum cost is $C_{\min} = 7201.90$ rupees.

So, if you were trying to help the trucker maximize his profit, you should advise him to turn down the job and stay home; he will lose less money that way.

The cost model in (2) is perfectly sensible, and is similar to many models encountered in real-world applications. However, in my opinion your profit model in (1) does not make economic or practical sense; I think it models the situation incorrectly. However, we can speak about that further off-line, perhaps in a PF "conversation".

 

[Krushnaraj Pandya]

You still have refused to answer two of my questions, even though I had good reasons to ask them. These are
(1) which model are you using---the original "profit" maximization in post #1 or the "cost" minimization proposed in post #5?
(2) what is your resulting minimum cost or maximum profit?

EDIT: I see that you at least hinted at the answer to question (1), although you were not very clear and explicit: you seemed to be saying that you were using model (1), shown below.

When I do it I have either
$$ \max\;\; 6 \left( \frac{400}{x}\right) - 400\left(12+ \frac x 6 \right), \; 35 \leq x \leq 60 \hspace{4ex} (1)$$
or
$$\min\;\; 6 \left( \frac{400}{x} \right) + 400\left(12+ \frac x 6 \right), \; 35 \leq x \leq 60 \hspace{4ex} (2)$$
Problem (1) is the profit maximization problem, while (2) is the cost minimization problem. Both have the solution $x = 35$ (km/hr). The maximum profit is $P_{\max} = - 7064.76$ rupees (a loss of 7064.76 rupees), while the minimum cost is $C_{\min} = 7201.90$ rupees.

So, if you were trying to help the trucker maximize his profit, you should advise him to turn down the job and stay home; he will lose less money that way.

The cost model in (2) is perfectly sensible, and is similar to many models encountered in real-world applications. However, in my opinion your profit model in (1) does not make economic or practical sense; I think it models the situation incorrectly. However, we can speak about that further off-line, perhaps in a PF "conversation".

Certainly! I'll message very soon. I apologize for replying so late- I have my JEE exams in a month and that keeps me quite busy but as soon as I'm done with a little but of my syllabus I'll get back and explain my work in detail- I do understand what you mean to say though, I appreciate you taking the time and interest to improve my knowledge. Thank you :D