Find maximum area of rectangle

A horizontal line satisfing $0 < y < \frac{4}{27}$ meets the graph in 3 points, $x = \alpha, \beta,$ , and $\gamma$ .

Say $\alpha$ is the largest, hence $\alpha > 1$ . Thus $x = \beta$ , and $x = \gamma$ constitute the vertical sides of rectangle.

Now, $x(x-1)^2 = \alpha(\alpha-1)^2$

Using vieta's formulae,

$\alpha + \beta + \gamma = 2$ ,

$\alpha \beta + \alpha \gamma + \gamma \beta = 1$

Multiplying first equation by $\alpha$ and subtracting second , we get:

$\beta \gamma = \alpha^2 - 2\alpha + 1$

Also, $\beta + \gamma = 2 - \alpha$

Hence, $|\beta - \gamma| = \sqrt{(\beta+ \gamma)^2 - 4 \beta \gamma} = \sqrt{4\alpha - 3\alpha^2}$ .

Hence, the sides of rectangle are $\alpha(\alpha-1)^2$ , and $\sqrt{4\alpha- 3\alpha^2}$

Thus , area is $A(\alpha) = \alpha(\alpha-1)^2\sqrt{4\alpha- 3\alpha^2}$

Maximise it to get $\boxed{A_{max} = \dfrac{7 \sqrt{7} - 10}{108 \sqrt{3}}}$

Doing this in a straightforward way, let $x=p$ be the left side of the rectangle. Then the other point is a root of the equation

$x{ (x-1) }^{ 2 }-p{ (p-1) }^{ 2 }=0$

But $(x-p)$ is already one of the factors, so dividing this factor out, we have the quadratic to solve

${ x }^{ 2 }+(p-2)x+{ (p-1) }^{ 2 }=0$

which the smaller of the roots $>p$ is

$x=\dfrac { 1 }{ 2 } (2-p-\sqrt { 4p-3{ p }^{ 2 } } )$

We then have the area of the rectangle to maximize

$(\dfrac { 1 }{ 2 } (2-p-\sqrt { 4p-3{ p }^{ 2 } } )-p)p{ (p-1) }^{ 2 }$

A bit of tedious calculus gets us the maximum

$\dfrac { 7\sqrt { 7 } -10 }{ 108\sqrt { 3 } }$

so that $a+b+c+d=128$