1996 AHSME Problems/Problem 25

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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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