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 Howdy, 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. Thanks! 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. -K
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