Split This Maximization Into Two Parts

I feel like I didn't really follow the problem title's suggestion, but anyway here's a calculus-based solution.

Let $f(x,y)=5-y^2-4y-x^2+xy+3x\,.$ We can maximize this function by finding where both partial derivatives of the function reach a critical point (and thus where the gradient of the function is zero). Taking our derivatives and setting them equal to zero,

$\displaystyle \frac{\partial f}{\partial y}= -2y-4+x = 0$

$\displaystyle \frac{\partial f}{\partial x}= -2x+y+3 = 0$

Solving this system of equations by substitution (or whatever method you prefer), we find that the point where the gradient of the function equals zero is $\displaystyle \left(\frac{2}{3},\, -\frac{5}{3} \right)\,.\,$ Substituting these values into our function $f(x,y)\,$ , we find the function's maximum value to be $\dfrac{28}{3}.\,$ So our final answer is $\,\boxed{31}\,.$

Here's the solution that I believe Finn is going for. The expression is equal to

$\frac{28}{3} - \frac{1}{4} (2x-y-3)^2 - \frac{1}{12} ( 3y + 5 ) ^ 2$

Hence, the maximum occurs when both square are 0, or when $y = - \frac{5}{3}, x = \frac 2 3$ .

Another way of expressing it is as

$\frac{28}{3} - \frac{1}{4} ( x - 2y -4)^ 2 - \frac{1}{12} (3x-2)^2.$

These expressions follow from understanding the classification of conic sections.

The maximum value of: $5 - y^{2} - 4y - x^{2} + xy + 3x = \frac{28}{3}$

where $x = \frac{2}{3}$ and $y = \frac{-5}{3}$

Thus, $a = 28$ and $b = 3$

So, the answer is: $a + b = 28 + 3 = \boxed{31}$

the maximum are 28/3 at (x,y)=(2/3;-5/3) so a+b=28+3=31