Visiting various fields of mathematics!

Rewriting by completing the square, it becomes $\large (x-15)^{2}+(y-20)^{2}=7^{2}$ .

Then I created a parametric pair based on the equation being a circle, let $\large x=15+7cos(t)$ and $\large y=20+7sin(t)$

Next let $\large f := \frac{y}{x} = \frac {20+7sin(t)}{15+7cos(t)}$ and derive for critical points, when f'(t)=0.

$\large f'= \frac{7 (20 sin(t) + 15 cos(t) + 7)}{(15 + 7 cos(t))^2}=0 \Rightarrow 20sin(t)+15cos(t)+7=0$

Then allow $\large a=tan(\frac{t}{2})$ so that $\large sin(t)=\frac{2a}{a^{2}+1}$ and $\large cos(t)=\frac{1-a^{2}}{a^{2}+1}$

$\large 20sin(t)+15cos(t)+7=0 \Rightarrow 20\frac{2a}{a^{2}+1}+15\frac{1-a^{2}}{a^{2}+1}+7=0 \Rightarrow \quad a=\frac{-1}{2}\lor\large a=\frac{11}{2}$

These $\large a$ values also mean the pairs for $\large (cos(t),sin(t))$ are $\large (\frac{3}{5},\frac{-4}{5})\land\large (\frac{-117}{125},\frac{44}{125})$

Moving forward this means our pairs for $\large (x,y)$ are $\large (\frac{96}{5},\frac{72}{5})\land\large (\frac{1056}{125},\frac{2808}{125})$

Which brings us to our two $\large f=\frac{y}{x}$ values of $\large \frac{72}{96}=\frac{3}{4}\land\large \frac{2808}{1056}=\frac{117}{44}$ .

The larger of which being the maximum $\frac{y}{x}$ is $\large \frac{117}{44}$ thus the required answer is $\large 117+44=161$

I loved every minute of this problem! bringing me way back to middle school and learning about conic sections! Thank you for the awesome problem.

geometric solution:

Since we want to maximize $\large \quad\frac{y}{x}\quad$ , the line $\large\quad y=kx\quad$ will be tangent to the circle

We can write a location for x and y, using the similar triangles in green

Thus we get the equation $\large\quad 15-7sin\theta=24cos\theta\quad$ and $\large\quad 20+7cos\theta=24sin\theta$

Let $\large\quad 24 cos\theta = x \iff cos\theta=\frac{x}{24}\quad$ and likewise $\large\quad 24sin\theta=y \iff sin\theta=\frac{y}{24}$

Now we can solve for x and y (from above):

$\begin{array}{lll} 5-7sin\theta=24cos\theta &\qquad& 20+7cos\theta=24sin\theta \\ \large \Rightarrow 15-\frac{7y}{24}=x &\qquad& \Rightarrow 20+\frac{7x}{24}=y \\ \Rightarrow 15-\frac{7(20+\frac{7x}{24})}{24}=x & \qquad \textbf{and}\qquad &\Rightarrow 20+\frac{7(15-\frac{7y}{24})}{24}=y \\\Rightarrow \frac{15\cdot 24^2-7\cdot 20\cdot 24}{24^2+7^2}=x &\qquad&\Rightarrow \frac{20\cdot 24^2 +7\cdot 15\cdot 24}{24^2+7^2}=y \end{array}$

The answer requires

$\large \frac{y}{x}=\frac{20\cdot 24^2 + 7\cdot 15\cdot 24}{15\cdot 24^2 -7\cdot 20\cdot 24}=\frac{4\cdot 24+ 7\cdot 3}{3\cdot 24 -7\cdot 4}=\frac{96+21}{72-28}=\frac{117}{44}$

I'm just going to explain the method I used. Solve for $y$ . Then divide the result by $x$ , to obtain an expression for $\frac{y}{x}$ that's only in terms of $x$ . Then set its derivative to zero. This results in a quadratic equation with two rational roots. Then substitute each root into the expression. One of them will produce the maximum.