An algebra problem by Filippo Olivetti

The given expression can be written as a sum of square of two terms.

$(x-k)^2 + (5y+4k)^2 = 1$

Since the sum of two square terms is 1 therefore if we assume $x-k = cos\theta$ then $5y+4k = sin\theta$

Therefore $4x +5y = 4(cos\theta +k) + (sin\theta -4k) = \boxed{4cos\theta + sin\theta}$

Since the maximum value of $(4cos\theta + sin\theta)$ is $\sqrt{17}$ , Hence the required answer for the maximum value of $4x+5y$ is $\boxed{\sqrt{17}}$

After having noticed that there are two squares, the equation can be rewrite as: $\large (x-k)^2+(5y+4k)^2=1$ Let $x-k = a$ and $5y+4k=b$ , we need to find the maximum value of $4x+5y = 4a+b$ . So, we know that: $\large \left\{ \begin{matrix} a^2+b^2=1\\ 4a+b=t\\ \end{matrix} \right.$ where $t$ is the highest possible. This can be easly solved by QM-AM inequality. How can we find the coefficient of the terms we will use? Let's consider a general case (all the letters are real variables): $\large ( \underbrace{ha, ha, \dots, ha}_{j \; times}, \underbrace{qb, qb, \dots, qb}_{g \; times})$ where we have taken $j$ $ha$ (so $j$ is clearly a positive integer) and $g$ $qb$ (where $g$ is a positive integer) So QM-AM is: $\large \sqrt{ \frac{jh^2 a^2+gq^2b^2}{j+g} }\ge \frac{jha+gqb}{j+g}$

We need to have specific coefficient in order to obtain $a^2+b^2$ and $4a+b$ . So $\large \left\{ \begin{matrix} jh^2=1\\ gq^2=1\\ jh=4\\ gq=1 \end{matrix} \right.$ Dividing the first equation with the second and the third with the fourth, we obtain $h=\frac{1}{4}, q=1, j=16,g=1$ . So the terms of QM-AM are $\large (\underbrace{\frac{a}{4}, \frac{a}{4}, \dots, \frac{a}{4}}_{16 \; times}, b)$ $\large \rightarrow \sqrt{ \frac{a^2+b^2}{17} }\ge \frac{4a+b}{17} \rightarrow \sqrt{ \frac{1}{17} }\ge \frac{4a+b}{17} \rightarrow {4a+b} \le \sqrt{17}$ . Therefore, the maximum value is $\boxed{\sqrt{17}}$ .

Notice the quadratic in k and impose condition that the discriminant should be non negative so that k can have real values

Cheating a little bit, one may let $k=0$ , which leads to $x^2=1-25y^2$ . We use this to minimize $4x+5y$ using substitution and derivatives. This is certainly no proof whatsoever, yet.