Optimization Part I

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Hckyplayer8

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Find two positive numbers with a sum of 100 and a product that is as large as possible.

Given information is

x + y = 100
x times y = large as possible
Domain = [0,100]

So we are dealing with a problem asking us to find the absolute max.

I'm not sure of what to do next. Regarding earlier sections dealing with absolute max/mins, the equation was given, the derivative was found and then it was set to 0.

Since the domain is fairly small, it's pretty easy to logic my way through this and figure out that the answer is 50 and 50. But what is the correct way to do this in regards to Calculus?
 
Identify what is it is you are trying to optimise. Here it is the product: P = xy.

Use the other piece of info given to write P in terms of one variable only. x + y =100 so y = ...…......

Therefore P = .................. (in terms of x only).

There's your function, so now proceed in the usual way.
 
Harry_the_cat said:
Identify what is it is you are trying to optimise. Here it is the product: P = xy.

Use the other piece of info given to write P in terms of one variable only. x + y =100 so y = ...…......

Therefore P = .................. (in terms of x only).

There's your function, so now proceed in the usual way.
Click to expand...

So y = 100 - x

So P= (100-x)(x)

So multiple that to get P = 100y - y2 ?
 
You changed from \(x\) to \(y\), but in this case it doesn't matter since the two variables may be interchanged (cyclic symmetry).
 
MarkFL said:
You changed from \(x\) to \(y\), but in this case it doesn't matter since the two variables may be interchanged (cyclic symmetry).
Click to expand...

Oops. I reversed them on paper as well.

So next I have to prove this is a maximum which should be as simple as the second derivative test.

P''(y) = 0 - 2 = -2

Since P'' is less than zero, then the function has a relative maximum at y = 50 and f(y) = 50.
 
Your objective function is:

[MATH]f(x,y)=xy[/MATH]
Subject to the constraint:

[MATH]g(x,y)=x+y-100=0[/MATH]
Assuming we aren't going to use Lagrange Multipliers, or utilize cyclic symmetry, we can use the constraint to express the objective function in one variable, either \(x\) or \(y\). We typically use \(x\), and so we need \(y\) in terms of \(x\), and we find from the constraint:

[MATH]y=100-x[/MATH]
And so our objective function is:

[MATH]f(x)=x(100-x)=100x-x^2[/MATH]
Hence, as you found:

[MATH]f'(x)=100-2x=2(50-x)=0[/MATH]
And so our critical value is:

[MATH]x=50[/MATH]
We could have found this using the axis of symmetry of a quadratic function as well, which will be midway between the roots. You used the second derivative test t identify whether the critical value is at a maximum or a minimum, but we could also observe that the first derivative is positive to the left of the critical value, and negative to the right. Either method is fine, and both show we have a maximum. THus we may state:

[MATH]f_{\max}=f(50)=50(100-50)=50^2=2500[/MATH]
 
You are trying to max p= 100x-x^2 and found a critic point of x= 50.
It is NOT necessary at all to find f" (50) or to even look at both sides of f' (x) on both sides of 50 (although this is NOT wrong and maybe good practice). The reason is that p(x) = 100x-x^2 is a negative parabola and ALWAYS has a min value and never has a max value (unless you want to call the max infinity). So the critical value x=50 must be a min.
As I said, it may be good practice for you to look at the f' (x) near 50 to determine if you have a max or min OR look at the value of f"(50) BUT you should also know about parabola!! There are enough exercises in your book that will require you to use your method or MarkFL's method but in my opinion this problem just needs you to know about parabola. Please study their properties.
 
Jomo said:
You are trying to max p= 100x-x^2 and found a critic point of x= 50.
It is NOT necessary at all to find f" (50) or to even look at both sides of f' (x) on both sides of 50 (although this is NOT wrong and maybe good practice). The reason is that p(x) = 100x-x^2 is a negative parabola and ALWAYS has a min value and never has a max value (unless you want to call the max infinity). So the critical value x=50 must be a min.
As I said, it may be good practice for you to look at the f' (x) near 50 to determine if you have a max or min OR look at the value of f"(50) BUT you should also know about parabola!! There are enough exercises in your book that will require you to use your method or MarkFL's method but in my opinion this problem just needs you to know about parabola. Please study their properties.
Click to expand...
Jomo, I think you have your MIN and MAX mixed up! In remembrance of Denis, I send you to the corner! :cry: (I think he'd find that funny. :ROFLMAO:)
 
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