Difference between revisions of "1996 AHSME Problems/Problem 25"
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+ | ==Solution== | ||
+ | |||
+ | The first equation is a circle, so let's find its center and radius: | ||
+ | <math>x^2 - 14x + y^2 - 6y = 6</math> | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
==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:
So we have a circle centered at with radius , and we want to find the max of .
The set of lines are all parallel, with slope . Increasing 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 will be tangent to the circle.
Imagine that this line hits the circle at point . The slope of the radius connecting the center of the circle, , to tangent point will be , since the radius is perpendicular to the tangent line.
So we have a point, , and a slope of that represents the slope of the radius to the tangent point. Let's start at the point . If we go units up and units right from , we would arrive at a point that's units away. But in reality we want to reach the tangent point, since the radius of the circle is .
Thus, , and we want to travel up and over from the point to reach our maximum. This means the maximum value of occurs at , which is
Plug in those values for and , and you get the maximum value of , which is option
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 |