Global Trends

Let $F$ denote the expression in question. By completing the square,

$\begin{array} { r c l } 5F &=& 25x^2 + 60xy + 60y^2 + 40x + 4k y + 35 \\ &=& (5x+6y + 4)^2 + [24y^2 + y(-48 + 5k) + 19 ] \\ \phantom0 \\ &=& (5x+6y + 4)^2 + 6 \Bigg [ 4y^2 + y\left(\frac{5k}6 - 8\right) + \frac{19}6 \Bigg ] \\ \phantom0 \\ &=& (5x+6y + 4)^2 + 6 \left(2y + \frac{5k}{24} - 2\right)^2 - \dfrac{5(5k^2 - 96k + 96)}{96} \end{array}$

Hence, $F$ is minimized when $5x + 6y + 4 = 2y + \frac{5k}{24} - 2 = 0$ and has a minimum value of $M:= -\dfrac{5k^2 - 96k + 96}{96} = -\dfrac{5}{96}k^2 + k - 1$ .

For $M$ to be an integer, it is sufficient for $5k^2$ to be divisible by 96. Since 5 is coprime to 96, then $k^2$ is divisible by 96.

Since $96 = 2^5 \times 3$ . Then $\min(k) = 2^{\lceil 5/2 \rceil} \times 3 = \boxed{24} .$

Upon substitution, $F$ has a global minimum of $-7$ when $x = 1, y = -\frac32$ .

---

Here's a calculus solution:

Let $F$ denote the expression in question. At its global minimum, the partial derivatives of $F$ must be 0. That is,

$\dfrac{\partial F}{\partial x} = 10x + 12y + 8 = 0 , \quad \quad \dfrac{\partial F}{\partial y} = 12x + 24y + k = 0$

which yield $(x,y) = (\tfrac k8 - 2, 1 - \tfrac{5k}{48})$

Constructing the Hessian matrix shows that $F$ is minimized at the coordinate stated above. Hence, $\min(F) = -\dfrac{5}{96}k^2 + k - 1$

Trial and error shows that the smallest positive integer value of $k$ such that $\min(F)$ is an integer is $\min(k) = \boxed{24}$ .