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April 6th, 2014, 12:24 AM  #1 
Senior Member Joined: Mar 2014 Posts: 112 Thanks: 8 
Given that $\displaystyle x^2\,+\,y^2\,=\,14x\,+\,6y\,+\,6$, what is the largest possible value that $\displaystyle 3x\,+\,4y$ can have? Obviously, Google will give us this and YouTube will give us this to determine the correct answer of 73. My solution uses this. If 3x + 4y = C then we need to determine the maximum value for C. 0 = x² + y²  14x  6y  6 = (x  7)²  7² + (y  3)²  3²  6 i.e. (x  7)² + (y  3)² = 8². Let b = C/4 or C = 4b where 0 = 3x/4 + y  b. $\displaystyle \frac{\frac{3\,\times\,7}{4}\,+\,3\,\,b}{\sqrt{\frac{9}{16}\,+\,1}}\,=\,8$ i.e. the distance from center (7, 3) to the point of tangency equals to the radius 8. $\displaystyle \frac{33}{4}\,\,b\,=\,10$ i.e. $\displaystyle b\,\,\frac{33}{4}\,=\,10$ so b = 10 + 33/4 = 40/4 + 33/4 = (40 + 33)/4 = 73/4 ≠ 7/4 (∵ b = 73/4 > 7/4). ∴ C = 4 × 73/4 = 73 ≠ 7 (∵ C = 73 > 7). Last edited by MarkFL; April 6th, 2014 at 12:58 AM. 
April 6th, 2014, 12:56 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs 
We could use Lagrange multipliers. We have the objective function: $\displaystyle f(x,y)=3x+4y$ Subject to the constraint: $\displaystyle g(x,y)=x^2+y^214x6y6=0$ We then obtain the system: $\displaystyle 3=\lambda(2x14)$ $\displaystyle 4=\lambda(2y6)$ which implies: $\displaystyle y=\frac{4}{3}(x7)+3$ Substituting for $y$ into the constraint, we obtain: $\displaystyle x^2+\left(\frac{4}{3}(x7)+3 \right)^214x6\left(\frac{4}{3}(x7)+3 \right)6=0$ Expand and distribute: $\displaystyle x^2+\frac{16}{9}(x7)^2+8(x7)+914x8(x7)186=0$ $\displaystyle \left(x^214x+49 \right)+\frac{16}{9}(x7)^2+918649=0$ $\displaystyle \frac{25}{9}(x7)^2=8^2$ $\displaystyle \frac{5}{3}(x7)=\pm8$ $\displaystyle x7=\pm\frac{24}{5}$ $\displaystyle x=7\pm\frac{24}{5}$ Hence: $\displaystyle y=\frac{4}{3}\left(\pm\frac{24}{5} \right)+3=3\pm\frac{96}{15}$ Checking the objective function's values at these critical points, we find: $\displaystyle f_{\max}=f\left(\frac{59}{5},\frac{47}{5} \right)=73$ $\displaystyle f_{\min}=f\left(\frac{11}{5},\frac{17}{5} \right)=7$ 
April 6th, 2014, 01:00 AM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs 
To the OP: Please do not post a thread and then remove the content. I have restored the original content so that the post I worked to produce is not "orphaned." When you create a thread and then gut it by removing the contents of the original post you devalue the thread and potentially waste the time of those who responded. 
April 6th, 2014, 01:02 AM  #4 
Senior Member Joined: Mar 2014 Posts: 112 Thanks: 8 
I noticed that MarkFL did reply to this one straightforward. In the Math Forums > Math Events, I posted there because I was strict about where I should post this, but again MarkFL is a moderator . . .

April 6th, 2014, 01:05 AM  #5 
Senior Member Joined: Mar 2014 Posts: 112 Thanks: 8 
I hope you won't feel the "wasted time" because I did quote your entire post to the other section as you can see. My question is, regardless of this section are you going to delete the other similar content at Math Forums > Math Events? I hope not.

April 6th, 2014, 01:07 AM  #6  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs  Quote:
I will move this thread there and delete the duplicate.  
April 6th, 2014, 01:12 AM  #7 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs  In order to avoid duplication, I did remove the second thread as there is no need for two copies of the same thread on a forum. I do appreciate that you made an effort to preserve my work. 
April 7th, 2014, 10:47 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 16,583 Thanks: 1199  By default, reports cause a message to be sent to all the moderators, so it might be simpler to check who's online, as there's usually a moderator online, and then send a PM to that moderator.
