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Finding Extreme Values for Polynomials

Date: 6 Jul 1995 11:16:32 -0400
From: Margie Salaz
Subject: Algebra question

Dear Dr. Math:

I am currently working on polynomial functions, finding the maximum and
minimum values.  I have been unable to come up with the right answer for the
following question:

Find the smallest possible value of the quantity X^2 + Y^2 under the
restriction that 2X + 3Y = 6.

The correct answer is: 36/13

Will you show me the process whereby this correct answer is derived?

Thank you very much for helping me with this problem.

Date: 6 Jul 1995 14:17:03 -0400
From: Dr. Ethan
Subject: Re: Algebra question

Good problem.

Here is how I would approach it.

Since we know that 2X + 3Y = 6 we can think of Y as dependent upon X.

Or Y = (6 - 2X)/3 So finding the Max of x^2 + y^2 is the same as finding the
max of x^2   + (  (6 - 2x)/3 )^2  which simplifies to 

4 - 8x/3 + (13x^2)/9

Now you can just find the maximum the way you normally would for a single
variable quadratic polynomial.

See if this helps if not right back and I will give you more.

Ethan Doctor On Call

From: Margie Salaz
Subject: Algebra question

Dear Dr. Math,

You helped me with this problem once before, but I'm still not coming out 
right.  (Ethan Magness was the Dr. on call when I requested help for this one.)

Here is the problem:  Find the smallest possible value of the quantity  x^2 
+ y^2 under the restriction that 2x + 3y = 6.

Here is what I did:  2x + 3y = 6
                                  x +3/2y = 3
                                  3/2y = 3-x
                                  y = 2/3(3-x)
                                   y = 2-2/3x
x^2 + y^2
x^2 + (2-2/3x)^2
x^2 + 4 - 8/3x + 4/9x^2
13x^2 - 24x + 36
13 (x^2 - 24/13x +(24/26)^2) +36
13 (x-12/13)^2 + 36 -1872/169 = (468/13-144/13) = (324/13)
13(x-12/13)^2 + 324/13
x = 12/13  and y = 324/13

Thus the smallest possible value of the quantity would by the y value of 
324/13.  However, the answer key in the back of the book gives the answer as 
36/13.  What am I doing wrong?? I have done this problem 100 times and I 
can't see what's wrong.


Frustrated in NM,

Margie Salaz
Cuba, NM

Date: 14 Jul 1995 11:12:22 -0400
From: Dr. Ken
Subject: Re: Minimum value

Hello there!

I'm happy to say that your error is just a simple one that's easily fixed.
I'll show it to you where you did it:

>Here is what I did:  2x + 3y =6
>                                  x +3/2y=3
>                                  3/2y = 3-x
>                                  y = 2/3(3-x)
>                                   y = 2-2/3x
>x^2 + y^2
>x^2 + (2-2/3x)^2
>x^2 + 4 - 8/3x + 4/9x^2

Right here.  You multiplied by 9 from this step to the next.  You'll notice
that your answer is nine times the answer given.  This is no coincidence.  

The reason you can't multiply by 9 here is that you're trying to find the
maximum value of the function, and if you multiply the value of the function
by 9, well, you'll get 9 times what you want.  However, by conincidence,
you'll get the right x value (if you understand why you get the right x
value, then by all means use this method, but it's a little dangerous), but
then to get the right y value you should substitute back in to your equation
2x + 3y = 6.  This is an equation that's true for all time.

The best way to do the problem would be to do it all without multiplying
through by anything.  It's a little more messy, but less confusing.

Associated Topics:
College Calculus

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