Max & Min Problem: Find Two Positive Numbers

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In summary, the problem is finding a maximum problem and trying to simplify the equation to find the values for the maximum. The first time I tried to do this I got the wrong answer and needed to use a common x value in the equation to get the correct answer. The second time I tried to do the first derivative but couldn't. If anyone could give me some direction I would really appreciate it.
  • #1
scorpa
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Hello Everyone,

I am having trouble with a maximum problem and I'm not quite sure where I am going wrong so I will type of the problem and what I have done so far.

Find two positive numbers whose sum is 18 and the product of the first number and the square of the other is a maximum.

Here is what I've done so far:

x + y = 18 ---> y = 18 - x

xy^2=P

P = x(18-x)^2
P = x(324 - 36x + x^2)
P= x^3 - 36x^2 + 324X

To find where there is a maximum I found the first derivative of the equation above:

dP/dx = 3x^2 - 72x + 324

This is where I'm stuck, I know I want to make the first derivative equal to zero so I can find the values for the maximum, and verify my answer using the second derivative, but the first derivative cannot be factored. I must be doing something terribly wrong. The first time I did it I got x = 18 and y = 0 which cannot be right, to get that answer I took out a common x value in the equation P equation, but when I did the second derivative test it showed that the answer was actually a minimum. If anyone could give me some direction here I would really appreciate it.
 
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  • #2
It doesn't need to be factorable, worst come to worst, you can use the quadratic formula.

Also dP/dx does simplify.

[tex] (18-x)^2 - 2x(18-x) = (18-x)(18-x-2x) = (18-x)(18-3x)[/tex]
 
  • #3
Ok turns out I'm mighty dumb...lol.

so dP/dx= (x-18)(x-6) = 0
therefore x can equal 18 or 6
when you take the second derivative the only number that equals a maximum is 6, so the two numbers are x = 6 and y = 12.

I think that is right now.
 
  • #4
so dP/dx= (x-18)(x-6) = 0

This doesn't equal [itex] 3x^2 - 72x + 324[/itex]
 
  • #5
Actually it does, since 3, 72, and 324 are all divisible by three the polynomial can be simplified to x^2 -24x + 108 which can be factored to be (x -18)(x -6)
 
  • #6
But you still have to tag a 3 on there.. expand your factors.. you get x^2-24x+108, which is 3 times less than the original polynomial.

When you solve for x you can eliminate it though, which is what I think ur doing.
 
  • #7
Oh right, my bad. Just the sort of thing I'd get on a students case about writing, I'm such a hypocrite. :(
 

Related to Max & Min Problem: Find Two Positive Numbers

1. What is a "Max & Min Problem"?

A "Max & Min Problem" is a type of mathematical problem that involves finding the maximum and minimum values of a given set of numbers or variables. In the case of "Find Two Positive Numbers", the problem involves finding the largest and smallest possible values for two positive numbers.

2. What is the general approach to solving a Max & Min Problem?

The general approach to solving a Max & Min Problem involves identifying the constraints or conditions of the problem, setting up an equation or system of equations, and then using mathematical techniques such as differentiation or substitution to find the maximum and minimum values.

3. How do you find the maximum and minimum values of two positive numbers?

To find the maximum and minimum values of two positive numbers, you can use the following steps:

1. Let the two numbers be represented by variables, such as x and y.

2. Write an equation or system of equations that represents the given conditions, for example, x + y = 10.

3. Use mathematical techniques such as differentiation or substitution to solve the equation(s).

4. Check the solutions to make sure they satisfy the given conditions, such as being positive numbers.

5. The largest solution is the maximum value, and the smallest solution is the minimum value.

4. What are some real-life applications of Max & Min Problems?

Max & Min Problems have many real-life applications, such as optimizing production costs in a manufacturing process, determining the maximum and minimum values of stock prices for investment decisions, and finding the optimal route for a delivery truck to minimize time and fuel costs.

5. Are there any limitations or assumptions when solving a Max & Min Problem?

Yes, there may be limitations or assumptions when solving a Max & Min Problem. Some common limitations include assuming a linear relationship between variables, ignoring external factors that may affect the outcome, or not considering all possible solutions. It is important to carefully analyze the problem and its constraints to ensure accurate results.

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