# 1996 AHSME Problem 25

#### the john

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 = xÂ² + yÂ² - 14x - 6y - 6 = (x - 7)Â² - 7Â² + (y - 3)Â² - 3Â² - 6 i.e. (x - 7)Â² + (y - 3)Â² = 8Â².
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 â‰  -7/4 (âˆµ b = 73/4 > -7/4).
âˆ´ C = 4 Ã— 73/4 = 73 â‰  -7 (âˆµ C = 73 > -7).

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#### MarkFL

We could use Lagrange multipliers. We have the objective function:

$$\displaystyle f(x,y)=3x+4y$$

Subject to the constraint:

$$\displaystyle g(x,y)=x^2+y^2-14x-6y-6=0$$

We then obtain the system:

$$\displaystyle 3=\lambda(2x-14)$$

$$\displaystyle 4=\lambda(2y-6)$$

which implies:

$$\displaystyle y=\frac{4}{3}(x-7)+3$$

Substituting for $y$ into the constraint, we obtain:

$$\displaystyle x^2+\left(\frac{4}{3}(x-7)+3 \right)^2-14x-6\left(\frac{4}{3}(x-7)+3 \right)-6=0$$

Expand and distribute:

$$\displaystyle x^2+\frac{16}{9}(x-7)^2+8(x-7)+9-14x-8(x-7)-18-6=0$$

$$\displaystyle \left(x^2-14x+49 \right)+\frac{16}{9}(x-7)^2+9-18-6-49=0$$

$$\displaystyle \frac{25}{9}(x-7)^2=8^2$$

$$\displaystyle \frac{5}{3}(x-7)=\pm8$$

$$\displaystyle x-7=\pm\frac{24}{5}$$

$$\displaystyle x=7\pm\frac{24}{5}$$

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:

$$\displaystyle f_{\max}=f\left(\frac{59}{5},\frac{47}{5} \right)=73$$

$$\displaystyle f_{\min}=f\left(\frac{11}{5},-\frac{17}{5} \right)=-7$$

#### MarkFL

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.

1 person

#### the john

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 . . .

#### the john

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.

#### MarkFL

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 . . .
In the future, if you wish for a thread to be moved, please use the "Report Post" function and ask that it be moved. This will bring it to the attention of the staff, who will then move it for you.

I will move this thread there and delete the duplicate.

#### MarkFL

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.
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.

#### skipjack

Forum Staff
In the future, if you wish for a thread to be moved, please use the "Report Post" function and ask that it be moved.
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.