1996 AHSME Problems/Problem 25

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,\[ x = 7 + \frac{3 \cdot 8}{5} = \frac{59}{5}, \qquad y = 3 + \frac{4 \cdot 8}{5} = \frac{47}{5}, \]or\[ x = 7 - \frac{3 \cdot 8}{5} = \frac{11}{5}, \qquad y = 3 - \frac{4 \cdot 8}{5} = - \frac{17}{5}. \] [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.

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

Page

Toolbox

Search