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

(→‎Solution 4 (Using Answer Choice + Calculus))
(→‎Solution 6 (polar coordinates))
 
(18 intermediate revisions by 9 users not shown)
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<cmath>(x-7)^2 + (y-3)^2 = 64.</cmath>
 
<cmath>(x-7)^2 + (y-3)^2 = 64.</cmath>
 
Applying Cauchy-Schwarz directly,
 
Applying Cauchy-Schwarz directly,
−
<cmath>64*25=(3^2+4^2)((x-7)^2 + (y-3)^2) \ge (3(x-7)+4(y-3))^2.</cmath>
+
<cmath>64\cdot25=(3^2+4^2)((x-7)^2 + (y-3)^2) \ge (3(x-7)+4(y-3))^2.</cmath>
 
<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>
Line 17: Line 17:
  
 
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]]:  
−
<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^2 - 14x + y^2 - 6y = 6</math>, so <cmath>(x-7)^2 + (y-3)^2 = (x^2- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9 = 64.</cmath>
  
 
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 32: Line 32:
  
 
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{(B)}</math>.
 +
==Solution 2B==
 +
Let the tangent point be <math>P</math>, and the tangent line's x-intercept be <math>Q</math>. Consider the horizontal line starting from center of circle (O) meeting the tangent line at K. Now triangle <math>OPK</math> is 3-4-5, <math>OP=8</math>, so <math>OK = \frac{5}{3}*8 = \frac{40}{3}</math>. Note that the horizontal distance from <math>O</math> to the origin is <math>7</math>, and the horizontal distance from K to Q is 4, (<math>\frac{4}{3}</math> of its y coordinate), so the x-intercept is <math>7+4+OK = 73/3</math>. The value of <math>3x+4y</math> is 73 at point <math>Q</math>. Note that this value is constant on the tangent line, so there is no need to calculate the coordinate of <math>P</math>. <math>\boxed{(B)}</math>.
  
 
==Solution 3==
 
==Solution 3==
Line 56: Line 58:
 
Now solve for <math>\frac{dy}{dx}</math> to obtain:
 
Now solve for <math>\frac{dy}{dx}</math> to obtain:
  
−
<math>\frac{dy}{dx} = \frac{x - 7}{y - 3}</math>
+
<math>\frac{dy}{dx} = -\frac{x - 7}{y - 3}</math>
  
−
Set the equation equal to zero to find the maximum occurs at <math>x = 0</math>
+
Set the equation equal to zero to find the maximum occurs at <math>x = 7</math>
  
 
Plug this back into the equation that we are trying to maximize and see that we are left with:  
 
Plug this back into the equation that we are trying to maximize and see that we are left with:  
 
<math>21 + 4y</math>.
 
<math>21 + 4y</math>.
  
−
The only answer choice that can be obtained from this equation is \textbf{<math>73</math>}
+
The only answer choice that can be obtained from this equation is <math>\bf{73}</math>
 +
 
 +
==Solution 5 (Lagrange Multipliers)==
 +
First, we move all the non-constant terms of the constraint to one side and assign it to the function <math>g(x,y)</math>:
 +
<cmath>g(x,y)=x^2+y^2-14x-6y.</cmath>
 +
Since we are trying to maximize <math>f(x,y)=3x+4y</math>, we need to solve for <math>x</math> and <math>y</math> in the system
 +
<cmath>\begin{cases}x^2+y^2-14x-6y=6,\\\nabla g(x,y)=\lambda\nabla f(x,y).\end{cases}</cmath>
 +
We have that
 +
<cmath>\begin{align*}\nabla f(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}3x+4y\\\dfrac{\partial}{\partial y}3x+4y\end{pmatrix}\\&=\begin{pmatrix}3\\4\end{pmatrix},\end{align*}</cmath>
 +
and
 +
<cmath>\begin{align*}\nabla g(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}x^2+y^2-14x-6y\\\dfrac{\partial}{\partial y}x^2+y^2-14x-6y\end{pmatrix}\\&=\begin{pmatrix}2x-14\\2y-6\end{pmatrix}\end{align*}.</cmath>
 +
To solve the original system, we can solve for <math>x</math> and <math>y</math> in terms of <math>\lambda</math> using our equations from the gradients, then substitute them into the first equation. We have that <math>x=\frac{3}{2}\lambda+7</math> and <math>y=2\lambda+3</math>. Substituting into the first equation, we have that
 +
<cmath>\begin{align*}\left(\frac{3}{2}\lambda+7\right)^2+(2\lambda+3)^2-14\left(\frac{3}{2}\lambda+7\right)-6(2\lambda+3)&=6\\\frac{25}{4}\lambda^2&=64\\\lambda&=\pm\frac{16}{5}\end{align*}</cmath>
 +
Using the solutions of <math>x</math> and <math>y</math> in terms of <math>\lambda</math> that we found earlier, we have that
 +
<cmath>3x+4y=\frac{25}{2}\lambda+33.</cmath>
 +
Because we are trying to maximize this function, we will use the positive solution for <math>\lambda</math>. Therefore, after substituting, we have that the largest value of <math>g(x,y)</math> that satisfies <math>f(x,y)=6</math> is <math>\boxed{\textbf{(B) }73}</math>.
 +
 
 +
~qianqian07
 +
 
 +
==Solution 6 (polar coordinates)==
 +
Completing the square gives the equation of a circle, <math>(x-7)^2+(y-3)^2=64.</math> Seeing that we would like to maximize <math>3x+4y,</math> we parameterize the circle using polar coordinates:
 +
<cmath>\begin{align*}
 +
x&=7+8\cos\theta&y&=3+8\sin\theta.
 +
\end{align*}</cmath>
 +
Then, we have <math>3x+4y=3(7+8\cos\theta)+4(3+8\sin\theta)=33+24\cos\theta+32\sin\theta.</math> Since <math>a\cos\theta+b\sin\theta\le\sqrt{a^2+b^2},</math> the desired answer is <math>33+\sqrt{24^2+32^2}=33+40=\boxed{73}.</math>
 +
- Ultroid999OCPN
 +
==Solution 6 (Alcumus)==
 +
The equation <math> x^2 + y^2 = 14x + 6y + 6</math> can be written\[
 +
(x-7)^2 + (y-3)^2 = 8^2,
 +
\]which defines a circle of radius 8 centered at <math>(7,3)</math>. If <math>
 +
k</math> is a possible value of <math>3x + 4y</math> for <math>(x,y)</math> on the circle, then the line <math>3x + 4y = k</math> must intersect the circle in at least one point. The largest value of <math>k</math> occurs when the line is tangent to the circle, and is therefore perpendicular to the radius at the point of tangency. Because the slope of the tangent line is <math>-3/4</math> the slope of the radius is <math>\ 4/3</math>. It follows that the point on the circle that yields the maximum value of <math>3x + 4y</math> is one of the two points of tangency,\[
 +
x = 7 + \frac{3 \cdot 8}{5} = \frac{59}{5}, \hspace{.3in} y = 3 +
 +
\frac{4 \cdot 8}{5} = \frac{47}{5},
 +
\]or\[
 +
x = 7 - \frac{3 \cdot 8}{5} = \frac{11}{5}, \hspace{.3in} y = 3 -
 +
\frac{4 \cdot 8}{5} = - \frac{17}{5}.
 +
\][asy]
 +
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8));
 +
draw((0,15)--origin--(15,0));
 +
dot("<math>(7,3)</math>",(7,3),S);
 +
draw(Circle((7,3),8));
 +
line a = line((0,9),(12,0));
 +
line b = line((0,12),(16,0));
 +
line c = line((0,18.3),(18.3*4/3,0));
 +
draw(a^^b^^c);
 +
[/asy]
 +
 
 +
The first point of tangency gives\[
 +
3x + 4y = 3 \cdot \frac{59}{5} + 4 \cdot \frac{47}{5} =
 +
\frac{177}{5}+\frac{188}{5}= 73,
 +
\]and the second one gives<cmath>\, 3x + 4y = \frac{33}{5}-\frac{68}{5}
 +
= -7.</cmath>Thus, <math>\boxed{73}</math> is the desired maximum, while <math>-7</math> is the minimum.
  
 
==See also==
 
==See also==
Line 70: Line 123:
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
−
[[Category:Quadratic Equation Problems]]
+
 
 
[[Category:Circle Problems]]
 
[[Category:Circle Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
−
<math>aopsswag</math>
 

Latest revision as of 16:51, 22 October 2023

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, \[64\cdot25=(3^2+4^2)((x-7)^2 + (y-3)^2) \ge (3(x-7)+4(y-3))^2.\] \[40 \ge 3x+4y-33\] \[3x+4y \le 73.\] 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 \[(x-7)^2 + (y-3)^2 = (x^2- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9 = 64.\]

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 2B

Let the tangent point be $P$, and the tangent line's x-intercept be $Q$. Consider the horizontal line starting from center of circle (O) meeting the tangent line at K. Now triangle $OPK$ is 3-4-5, $OP=8$, so $OK = \frac{5}{3}*8 = \frac{40}{3}$. Note that the horizontal distance from $O$ to the origin is $7$, and the horizontal distance from K to Q is 4, ($\frac{4}{3}$ of its y coordinate), so the x-intercept is $7+4+OK = 73/3$. The value of $3x+4y$ is 73 at point $Q$. Note that this value is constant on the tangent line, so there is no need to calculate the coordinate of $P$. $\boxed{(B)}$.

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 $\boxed{(B)}$.


Solution 4 (Using Answer Choice + Calculus)

Implicitly differentiating the given equation with respect to $x$ yields:

$2x + 2y\frac{dy}{dx} = 14 + 6\frac{dy}{dx}$

Now solve for $\frac{dy}{dx}$ to obtain:

$\frac{dy}{dx} = -\frac{x - 7}{y - 3}$

Set the equation equal to zero to find the maximum occurs at $x = 7$

Plug this back into the equation that we are trying to maximize and see that we are left with: $21 + 4y$.

The only answer choice that can be obtained from this equation is $\bf{73}$

Solution 5 (Lagrange Multipliers)

First, we move all the non-constant terms of the constraint to one side and assign it to the function $g(x,y)$: \[g(x,y)=x^2+y^2-14x-6y.\] Since we are trying to maximize $f(x,y)=3x+4y$, we need to solve for $x$ and $y$ in the system \[\begin{cases}x^2+y^2-14x-6y=6,\\\nabla g(x,y)=\lambda\nabla f(x,y).\end{cases}\] We have that \begin{align*}\nabla f(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}3x+4y\\\dfrac{\partial}{\partial y}3x+4y\end{pmatrix}\\&=\begin{pmatrix}3\\4\end{pmatrix},\end{align*} and \begin{align*}\nabla g(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}x^2+y^2-14x-6y\\\dfrac{\partial}{\partial y}x^2+y^2-14x-6y\end{pmatrix}\\&=\begin{pmatrix}2x-14\\2y-6\end{pmatrix}\end{align*}. To solve the original system, we can solve for $x$ and $y$ in terms of $\lambda$ using our equations from the gradients, then substitute them into the first equation. We have that $x=\frac{3}{2}\lambda+7$ and $y=2\lambda+3$. Substituting into the first equation, we have that \begin{align*}\left(\frac{3}{2}\lambda+7\right)^2+(2\lambda+3)^2-14\left(\frac{3}{2}\lambda+7\right)-6(2\lambda+3)&=6\\\frac{25}{4}\lambda^2&=64\\\lambda&=\pm\frac{16}{5}\end{align*} Using the solutions of $x$ and $y$ in terms of $\lambda$ that we found earlier, we have that \[3x+4y=\frac{25}{2}\lambda+33.\] Because we are trying to maximize this function, we will use the positive solution for $\lambda$. Therefore, after substituting, we have that the largest value of $g(x,y)$ that satisfies $f(x,y)=6$ is $\boxed{\textbf{(B) }73}$.

~qianqian07

Solution 6 (polar coordinates)

Completing the square gives the equation of a circle, $(x-7)^2+(y-3)^2=64.$ Seeing that we would like to maximize $3x+4y,$ we parameterize the circle using polar coordinates: \begin{align*} x&=7+8\cos\theta&y&=3+8\sin\theta. \end{align*} Then, we have $3x+4y=3(7+8\cos\theta)+4(3+8\sin\theta)=33+24\cos\theta+32\sin\theta.$ Since $a\cos\theta+b\sin\theta\le\sqrt{a^2+b^2},$ the desired answer is $33+\sqrt{24^2+32^2}=33+40=\boxed{73}.$ - Ultroid999OCPN

Solution 6 (Alcumus)

The equation $x^2 + y^2 = 14x + 6y + 6$ can be written\[ (x-7)^2 + (y-3)^2 = 8^2, \]which defines a circle of radius 8 centered at $(7,3)$. If $k$ is a possible value of $3x + 4y$ for $(x,y)$ on the circle, then the line $3x + 4y = k$ must intersect the circle in at least one point. The largest value of $k$ occurs when the line is tangent to the circle, and is therefore perpendicular to the radius at the point of tangency. Because the slope of the tangent line is $-3/4$ the slope of the radius is $\ 4/3$. It follows that the point on the circle that yields the maximum value of $3x + 4y$ is one of the two points of tangency,\[ x = 7 + \frac{3 \cdot 8}{5} = \frac{59}{5}, \hspace{.3in} y = 3 + \frac{4 \cdot 8}{5} = \frac{47}{5}, \]or\[ x = 7 - \frac{3 \cdot 8}{5} = \frac{11}{5}, \hspace{.3in} y = 3 - \frac{4 \cdot 8}{5} = - \frac{17}{5}. \][asy] import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); draw((0,15)--origin--(15,0)); dot("$(7,3)$",(7,3),S); draw(Circle((7,3),8)); line a = line((0,9),(12,0)); line b = line((0,12),(16,0)); line c = line((0,18.3),(18.3*4/3,0)); draw(a^^b^^c); [/asy]

The first point of tangency gives\[ 3x + 4y = 3 \cdot \frac{59}{5} + 4 \cdot \frac{47}{5} = \frac{177}{5}+\frac{188}{5}= 73, \]and the second one gives\[\, 3x + 4y = \frac{33}{5}-\frac{68}{5} = -7.\]Thus, $\boxed{73}$ is the desired maximum, while $-7$ is the minimum.

See also

1996 AHSME (Problems • Answer Key • Resources)
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Problem 26
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