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

(→‎Solution 1)
(Added Solution 3)
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<cmath> 40 \ge 3x+4y-33 </cmath>
 
<cmath> 40 \ge 3x+4y-33 </cmath>
 
<cmath>3x+4y \le 73.</cmath>
 
<cmath>3x+4y \le 73.</cmath>
−
Thus our answer is <math>\text{B}</math>
+
Thus our answer is <math>\boxed{(B)}</math>.
  
−
==Solution 2==
+
==Solution 2 (Geometric)==
  
 
The first equation is a [[circle]], so we find its center and [[radius]] by [[completing the square]]:  
 
The first equation is a [[circle]], so we find its center and [[radius]] by [[completing the square]]:  
Line 31: Line 31:
 
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>
 
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{(\text{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{(B)}</math>.
 +
 
 +
==Solution 3==
 +
 
 +
Let <math>z = 3x + 4y</math>. Solving for <math>y</math>, we get <math>y = (z - 3x)/4</math>. Substituting into the given equation, we get
 +
<cmath>x^2 + \left( \frac{z - 3x}{4} \right)^2 = 14x + 6 \cdot \frac{z - 3x}{4} + 6,</cmath>
 +
which simplifies to
 +
<cmath>25x^2 - (6z + 152)x + (z^2 - 24z - 96) = 0.</cmath>
 +
 
 +
This quadratic equation has real roots in <math>x</math> if and only if its discriminant is nonnegative, so
 +
<cmath>(6z + 152)^2 - 4 \cdot 25 \cdot (z^2 - 24z - 96) \ge 0,</cmath>
 +
which simplifies to
 +
<cmath>-64z^2 + 4224z + 32704 \ge 0,</cmath>
 +
which can be factored as
 +
<cmath>-64(z + 7)(z - 73) \ge 0.</cmath>
 +
The largest value of <math>z</math> that satisfies this inequality is <math>\boxed{73}</math>, which is <math>\boxed{(B)}</math>.
  
 
==See also==
 
==See also==

Revision as of 17:59, 19 June 2016

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 1

Complete the square to get \[(x-7)^2 + (y-3)^2 = 64.\] Applying Cauchy-Schwarz directly, \[64*25=(3^2+4^2)((x-7)^2 + (y-3)^2) \ge (3(x-7)+4(y-3))^2.\] \[40 \ge 3x+4y-33\] \[3x+4y \le 73.\] Thus our answer is $\boxed{(B)}$.

Solution 2 (Geometric)

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 $\boxed{(B)}$.

Solution 3

Let $z = 3x + 4y$. Solving for $y$, we get $y = (z - 3x)/4$. Substituting into the given equation, we get \[x^2 + \left( \frac{z - 3x}{4} \right)^2 = 14x + 6 \cdot \frac{z - 3x}{4} + 6,\] which simplifies to \[25x^2 - (6z + 152)x + (z^2 - 24z - 96) = 0.\]

This quadratic equation has real roots in $x$ if and only if its discriminant is nonnegative, so \[(6z + 152)^2 - 4 \cdot 25 \cdot (z^2 - 24z - 96) \ge 0,\] which simplifies to \[-64z^2 + 4224z + 32704 \ge 0,\] which can be factored as \[-64(z + 7)(z - 73) \ge 0.\] The largest value of $z$ that satisfies this inequality is $\boxed{73}$, which is $\boxed{(B)}$.

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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