Maximizing $\frac{y+1}{x+2}$ when $(x-3)^2 + (y-3)^2 = 6$

Question

Suppose $x$ and $y$ are real numbers such that $(x-3)^2 + (y-3)^2 = 6.$ Than, maximize $\frac{y+1}{x+2}.$

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I do in fact realize that this is a double post, but it's a 5 year old question and I don't feel as if it is appropriate to bump it. I did as the original post hinted towards, setting $k = \frac{y+1}{x+2}$ and than writing the given equation as $$(x-3)^2 + (k(x+2) - 4)^2 = 6.$$ I than proceeded to expand and simplify, which gave me $$x^2(k^2 + 1) + x(4k^2 - 8k - 6) + (4k^2 - 16k + 19).$$ However, I am unsure where to go from here. Should I use the discriminant now?

Answer 1

Going from what you have (and avoiding calculus):

$(k^2 + 1)x^2 + (4k^2 -4k-3)x + (k^2 - 16k + 19)= 0$

If we plug this into the quadratic formula:

$x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$

When $b^2 - 4ac < 0$ the formula breaks.

Setting $b^2 - 4ac = 0$ will give us the extreme values of $k.$

$(4k^2 - 8k - 6)^2 - 4(k^2+1)(4k^2 - 16k + 19) \ge 0\\ (16k^4 - 64k^3 + (-48+64)k^2 + 96k + 36) - (16k^4-64k^3 + 4(19+4)k^2 -64k + 76)\ge 0\\ -76k^2 + 160k - 40 \ge 0\\ 19k^2 - 40k + 10 \le 0\\ $

And now use the quadratic formula again to find the range of $k.$

$\frac {20 - \sqrt {210}}{19}\le k \le \frac {20 + \sqrt {210}}{19} $

Answer 2

Hint: Alternatively we can utilise trigonometry

WLOG $x-3=\sqrt6\cos2t,y-3=\sqrt6\sin2t$

We need to maximize $$F=\dfrac{\sqrt6\sin2t +4}{\sqrt6\cos2t+5}$$

Now use https://en.m.wikipedia.org/wiki/Weierstrass_substitution#The_substitution to form a quadratic equation in $\tan t$

As $\tan t$ is real, the discriminant must be $\ge0$

Answer 3

The function and the constraint are $$ f(x,y) = \frac{y+1}{x+2}\\ \quad g(x,y) = (x-3)^2 + (y-3)^2 - 6 = 0 $$ using Lagrange multipliers $$ \nabla f = (-\frac{y+1}{(x+2)^2}, \frac{1}{x+2}), \quad \nabla g = (2x,2y) $$ We get the equations $$ \nabla f - \lambda \nabla g = 0\\ g(x,y) = 0 $$ in $\lambda, x$, and $y$. The first (vector) equation is the two equations $$ -\frac{y+1}{(x+2)^2} - \lambda 2x = 0 \tag{1} $$ $$ \frac{1}{x+2} - \lambda 2y = 0 \tag{2} $$ Mathematica gives me the solution set $$ x = \frac{1}{41}(93-4\sqrt{210}),\quad y = \frac{1}{41}(99+5\sqrt{210}),\quad \lambda = \frac{8400+971\sqrt{210}}{151620} \\ x = \frac{1}{41}(93+4\sqrt{210}),\quad y = \frac{1}{41}(99-5\sqrt{210}),\quad \lambda = \frac{8400-971\sqrt{210}}{151620} \\ $$

Answer 4

A bit of geometry.

Introduce new coordinates X,Y:

$Y := y+1;$ $X := x+2;$

1)$Y=CX;$

2)$(X-5)^2 + (Y-4)^2=6;$

Need to find the slope $C$ of the line $Y=CX$ s.t. it is tangent to the circle 2).

3)Distance of tangent line $Y=CX$ to the centre of circle $(5,4):$

$\dfrac{|-5C+4|}{\sqrt{1+C^2}}=√6;$

$(-5C+4)^2=6(1+C^2);$

A quadratic equation for $C$. Choose the larger solution.