Area of rectangles

Source

This can be represented through the areas of a rectangle.

We have to maximize the blue area ie $\text{max}(ab + bc + cd) - \text{min}(ad)$

The minimum area of $ad$ can be $1$ , when $a = 1$ and $d = 1$ .

The expression now becomes $b + bc + c = b + c(b+1)$

Also $\begin{aligned}b + c &= 61 \\ \implies b &= 61 - c \end{aligned}$ Using this we get $61 - c + c(62 - c) = -c^2 + 61c + 61 = f$ We know that the vertex of a parabola $\alpha x^2 + \beta x + c$ is given by $x = \dfrac{-\beta}{2a}$ $\implies c = 30.5$ Using nearest integer values

When $c = 30, f = 991$

Also when $c = 31, f = 991$

So maximum value of $ab + bc + cd = \boxed{991}$

$ab+bc+cd=(ab+bc+cd+ad)-ad=A-ad$

$A=ab+bc+cd+ad=(a+c)(b+d)$

product of two integers $a+c$ and $b+d$ , that sum up to a fixed integer is maximised when the two are the closest. Due to symmetry, we may take $a+c=31$ and $b+d=32$ . Therefore, $31 \times 32 =992$ is the maximum that $A$ achieves. Knowing that $a$ and $d$ are nonnegative integers, we take them to be $a=d=1$ , so that $ad$ is minimised and, consequently, $A-ad$ would be maximised in theory. We just need to find $c$ and $b$ , for which $(a+c)=(1+c)=31$ and $(b+d)=(b+1)=32$ . For $(a,b,c,d)=(1,31,30,1)$ the maximum $991$ is achieved.