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

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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