Difference between revisions of "1996 AHSME Problems/Problem 25"

(Created page with "==See also== {{AHSME box|year=1996|num-b=24|num-a=26}}")
 
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==Solution==
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The first equation is a circle, so let's find its center and radius:
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<math>x^2 - 14x + y^2 - 6y = 6</math>
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<math>(x- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9</math>
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<math>(x-7)^2 + (y-3)^2 = 64</math>
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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>.
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The set of lines <math>3x + 4y = A</math> are all parallel, with slope <math>-\frac{3}{4}</math>.  Increasing <math>A</math> shifts the lines up and/or to the right.
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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 <math>3x + 4y = A</math> will be tangent to the circle.
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Imagine that this line hits the circle at point <math>(a,b)</math>.  The slope of the radius connecting the center of the circle, <math>(7,3)</math>, to tangent point <math>(a,b)</math> will be <math>\frac{4}{3}</math>, since the radius is perpendicular to the tangent line.
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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>
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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>
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==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=24|num-a=26}}
 
{{AHSME box|year=1996|num-b=24|num-a=26}}

Revision as of 15:06, 20 August 2011

Solution

The first equation is a circle, so let's find its center and radius: $x^2 - 14x + y^2 - 6y = 6$

$(x- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9$

$(x-7)^2 + (y-3)^2 = 64$

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 $(7 +3\cdot \frac{8}{5}, 3 + 4\cdot \frac{8}{5})$, which is $(\frac{59}{5}, \frac{47}{5})$

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 $\boxed{B}$

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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