Finding Max and Min Application of Derivatives

In summary, the company can manufacture up to 276 grade A tires and 553 grade B tires a day where the profit is twice as much for an A tire as it is for a B tire. The most profitable number of each kind to manufacture is 276 grade A and 553 grade B tires.
  • #1
LadiesMan
96
0
Finding Max and Min "Application of Derivatives"

Hello There!

The following question I need help in and it would be appreciated. Thank you very much!
Manufacturing Tires Your company can manufacture x hundred grade A tires and y hundred grade B tires a day where 0 (greater than or equal to) x (less than or equal to) 4 and

y = (40-10x)/(5-x)

Your profit on a grade A tire is twice your profit on a grade B tire. What is the most profitable number of each kind to make?

Thank for your time!
 
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  • #2
Start by writing an expression for your total profit.

2 side point.
1. If you read the guidelines you agreed to when you registered you will know that we require you to do your own work.

2. We have forums specifically for homework. This thread has been moved there.
 
  • #3
alright, sry about that, I'm just new here!
 
  • #4
At a stationary value, the first derivative is zero. At a max value,y''=-ve and at a min value,y''=+ve
 
  • #5
i found there to be no critical points if I just took the derivative, but I'm not sure if I was correct.
 
  • #6
What did you take the derivitive of? Please show us.
 
  • #7
i just took the derivative of the equation given, as to finding the max when the derivative is set to zero.
 
  • #8
The given expression is NOT the profit. You need to think about the profit. That will be the number of items sold times the profit for each. Can you write an expression for the total profit? How about the profit for product A?
 
  • #9
Since your profit on an A tire is twice as much as the B tire then P = 2A + B. From what is given above, we know that A is the same thing as 100x. We also know that B is the same thing as 100y.You can go ahead and sub those into the profit equation from above. From here you will want everything to be one variable. You already have a nice relationship between x and y, so another simple substitution there should suffice. Once you do that, use that equation to look for extrema.
 
  • #10
wow thanks, i didn't see it from that point
 
  • #11
So this is how i did it, P = 2(100x) + 100y

and then i subbed y = (40 - 10x)/(5-x) for y in the profit equation, P = 2(100x) + 100y

I think I am misunderstanding what your trying to tell me

Then i took the derivative and got an "ugly" number whereas should be a whole number for x
 
  • #12
One of the things we are trying to tell you, that you are apparently misunderstanding, is "show us what you did". Okay, you substituted y= (40-10x)/(5-x) into P= 2(100x)+ 100y. What did you get? When you differentiated that what did you get?
 
  • #13
once i subbed it into the the profit equation i took the derivative and got x to be

(10 (+ or minus) squareroot 80)/ 2

The x should be a whole number according to the answer.

276 grade A and 553 grade B tires
 
  • #14
anyone got any ideas?
 
  • #15
sry about that i madea mistake in finding the derivative:

My answer when i took the derivative is x^3 - 10x^2+25x - 5

When i tried to get the x's I can't get what is the answer. Did i make a mistake
 
  • #16
If you will not answer the questions I asked, I cannot help you.
 
  • #17
what do u mean, do u want me to show it step by step?
 
  • #18
nevermind i got the answer! Thanks for all your help!
 

1. What is the concept of "Finding Max and Min Application of Derivatives"?

The concept of "Finding Max and Min Application of Derivatives" involves using derivatives, which are a mathematical tool that represents the rate of change of a function, to find the maximum and minimum values of a given function. This can be useful in various real-world applications, such as optimization problems, where finding the maximum or minimum value can help determine the most efficient solution.

2. How do you find the maximum or minimum value of a function using derivatives?

To find the maximum or minimum value of a function using derivatives, you first need to find the critical points of the function, which are the points where the derivative equals zero. Then, you can use the first or second derivative test to determine whether these points are maximum or minimum values. If the second derivative is positive, the critical point is a minimum value, and if the second derivative is negative, it is a maximum value.

3. What is the first derivative test?

The first derivative test is a method used to determine whether a critical point is a maximum or minimum value. It involves evaluating the first derivative of the function at the critical point. If the first derivative is positive, the critical point is a minimum value, and if it is negative, the critical point is a maximum value.

4. What is the second derivative test?

The second derivative test is a method used to determine whether a critical point is a maximum or minimum value. It involves evaluating the second derivative of the function at the critical point. If the second derivative is positive, the critical point is a minimum value, and if it is negative, the critical point is a maximum value. If the second derivative is zero, the test is inconclusive, and another method, such as the first derivative test, must be used.

5. In what real-world applications is "Finding Max and Min Application of Derivatives" useful?

"Finding Max and Min Application of Derivatives" can be useful in various real-world applications, such as maximizing profits in business, minimizing costs in manufacturing, optimizing resource allocation in engineering, and finding the most efficient path in transportation. It can also be used in fields such as economics, physics, and biology, to name a few.

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