Maximizing $f(x,y) = (a + x)(b + y)$ under the constraint $d=x+y$, where $a$, $b$, $d$ are known. 
Find the maximum of the function $f(x,y) = (a + x)(b + y)$ under the constraint $d = x + y$, where $a$, $b$, $d$ are known.

It seems "obvious" to me that you'd want to split it between the two so that both sides of the product are as equal as possible but no idea how to prove the result.
Sorry I don't know what kind of math this is so I tagged a few.

Since you are not acquainted with the trick mentioned in the comment, I elaborate on this a bit.
You surely know the binomial formula


*

*$(u+v)^2 = u^2+2uv+v^2$
From this it follows
$$(u+v)^2-(u-v)^2 = 4uv \Rightarrow uv =\frac 14((u+v)^2-(u-v)^2)$$
Now, you plug into this formula 


*

*$u= a+x$ and $v= b+y$
and use $d=x+y$. So, you get
$$(a+x)(b+y)=\frac 14((a+b+d)^2-(a-b-d+2x)^2)$$
Obviously, the square expression we subtract on the RHS has a minimal value of $0$.
So, the maximal value you are looking for is
$$\frac 14(a+b+d)^2$$
