Answer

Notice that the expression we are asked to maximise is almost the full multiplication of $a+c$ with $b+d$. Then we can write $ab + bc + cd = (a+c)(b+d)-ad$, and we can express $(a+c) = 63 - (b+d)$. If we call $t = a+c$, we are asked to maximise $t(63-t)-ad$. Now the first part is a quadratic equation - we maximise it by taking the derivative and setting to 0: $63-2t=0$, so $t=31.5$ maximises the expression. This means that $a+c=31.5$ and therefore $b+d=31.5$ too for them to sum up to 63. This yields $31.5^2$ for the maximum value for the first part.

Now to maximise the full expression, we must also minimise $ad$. Since we are told all of them are positive intergers (or 0), the minimum we can reach for $ad$ is 0, by setting one of them to 0. Therefore, the maximum we can reach for the entire expression is $31.5^2=992.25$