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Difference between revisions of "1996 AHSME Problems/Problem 25"
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==Solution==
 
==Solution==
  
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The first equation is a circle, so let's find its center and radius:
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The first equation is a circle, so we find its center and radius by [[completing the square]]:  
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<math>x^2 - 14x + y^2 - 6y = 6</math>
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<math>x^2 - 14x + y^2 - 6y = 6</math>, so <cmath>(x-7)^2 + (y-3)^2 = (x- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9 = 64.</cmath>
−
 
 
−
<math>(x- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9</math>
 
−
 
 
−
<math>(x-7)^2 + (y-3)^2 = 64</math>
 
  
 
So we have a circle centered at <math>(7,3)</math> with radius <math>8</math>, and we want to find the max of <math>3x + 4y</math>.
 
So we have a circle centered at <math>(7,3)</math> with radius <math>8</math>, and we want to find the max of <math>3x + 4y</math>.
Line 24: Line 20:
 
So we have a point, <math>(7,3)</math>, and a slope of <math>\frac{4}{3}</math> that represents the slope of the radius to the tangent point.  Let's start at the point <math>(7,3)</math>.  If we go <math>4k</math> units up and <math>3k</math> units right from <math>(7,3)</math>, we would arrive at a point that's <math>5k</math> units away.  But in reality we want <math>5k = 8</math> to reach the tangent point, since the radius of the circle is <math>8</math>.
 
So we have a point, <math>(7,3)</math>, and a slope of <math>\frac{4}{3}</math> that represents the slope of the radius to the tangent point.  Let's start at the point <math>(7,3)</math>.  If we go <math>4k</math> units up and <math>3k</math> units right from <math>(7,3)</math>, we would arrive at a point that's <math>5k</math> units away.  But in reality we want <math>5k = 8</math> to reach the tangent point, since the radius of the circle is <math>8</math>.
  
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Thus, <math>k = \frac{8}{5}</math>, and we want to travel <math>4\cdot \frac{8}{5}</math> up and <math>3\cdot \frac{8}{5}</math> over from the point <math>(7,3)</math> to reach our maximum.  This means the maximum value of <math>3x + 4y</math> occurs at <math>(7 +3\cdot \frac{8}{5}, 3 + 4\cdot \frac{8}{5})</math>, which is <math>(\frac{59}{5}, \frac{47}{5})</math>
+
Thus, <math>k = \frac{8}{5}</math>, and we want to travel <math>4\cdot \frac{8}{5}</math> up and <math>3\cdot \frac{8}{5}</math> over from the point <math>(7,3)</math> to reach our maximum.  This means the maximum value of <math>3x + 4y</math> occurs at <math>\left(7 +3\cdot \frac{8}{5}, 3 + 4\cdot \frac{8}{5}\right)</math>, which is <math>\left(\frac{59}{5}, \frac{47}{5}\right).</math>
  
−
Plug in those values for <math>x</math> and <math>y</math>, and you get the maximum value of <math>3x + 4y = 3\cdot\frac{59}{5} + 4\cdot\frac{47}{5} = \boxed{73}</math>, which is option <math>\boxed{B}</math>
+
Plug in those values for <math>x</math> and <math>y</math>, and you get the maximum value of <math>3x + 4y = 3\cdot\frac{59}{5} + 4\cdot\frac{47}{5} = \boxed{73}</math>, which is option <math>\boxed{(\text{B})}</math>
  
 
==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=24|num-a=26}}
 
{{AHSME box|year=1996|num-b=24|num-a=26}}
 +
 +
[[Category:Introductory Geometry Problems]]
 +
[[Category:Introductory Algebra Problems]]
 +
[[Category:Quadratic Equation Problems]]
 +
[[Category:Circle Problems]]

Revision as of 15:17, 20 August 2011

Problem

Given that $x^2 + y^2 = 14x + 6y + 6$, what is the largest possible value that $3x + 4y$ can have?

$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$

Solution

The first equation is a circle, so we find its center and radius by completing the square: $x^2 - 14x + y^2 - 6y = 6$, so \[(x-7)^2 + (y-3)^2 = (x- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9 = 64.\]

So we have a circle centered at $(7,3)$ with radius $8$, and we want to find the max of $3x + 4y$.

The set of lines $3x + 4y = A$ are all parallel, with slope $-\frac{3}{4}$. Increasing $A$ shifts the lines up and/or to the right.

We want to shift this line up high enough that it's tangent to the circle...but not so high that it misses the circle altogether. This means $3x + 4y = A$ will be tangent to the circle.

Imagine that this line hits the circle at point $(a,b)$. The slope of the radius connecting the center of the circle, $(7,3)$, to tangent point $(a,b)$ will be $\frac{4}{3}$, since the radius is perpendicular to the tangent line.

So we have a point, $(7,3)$, and a slope of $\frac{4}{3}$ that represents the slope of the radius to the tangent point. Let's start at the point $(7,3)$. If we go $4k$ units up and $3k$ units right from $(7,3)$, we would arrive at a point that's $5k$ units away. But in reality we want $5k = 8$ to reach the tangent point, since the radius of the circle is $8$.

Thus, $k = \frac{8}{5}$, and we want to travel $4\cdot \frac{8}{5}$ up and $3\cdot \frac{8}{5}$ over from the point $(7,3)$ to reach our maximum. This means the maximum value of $3x + 4y$ occurs at $\left(7 +3\cdot \frac{8}{5}, 3 + 4\cdot \frac{8}{5}\right)$, which is $\left(\frac{59}{5}, \frac{47}{5}\right).$

Plug in those values for $x$ and $y$, and you get the maximum value of $3x + 4y = 3\cdot\frac{59}{5} + 4\cdot\frac{47}{5} = \boxed{73}$, which is option $\boxed{(\text{B})}$

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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