The maximum possible area bounded by the parabola $y = x^2 + x + 10$ and a chord of the parabola of length $1$ is? The maximum possible area bounded by the parabola $y = x^2 + x + 10$ and a chord of the parabola of length $1$ is?
$(y-39/4)=(x+1/2)^2$, Vertex: $(-1/2, 3/4)$
How do I find the equation of the chord whose length is $1$?
 A: Such a chord would intersection the parabola at points $(x_1, y_1)$ and $(x_2, y_2)$, where $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = 1$. Notice that if you square this expression, you get $(y_2 - y_1)^{2} = 1 - (x_2 - x_1)^2$. Here, you'd have $y_2 = x^{2}_{2} + x_2 + 10$ and $y_1 = x^{2}_{1} + x_1 + 10$. 
A: Let the equation of the secant line be $y=mx+b$. Combining this with the equation of the parabola produces the quadratic equation $$x^2+(1-m)x+(10-b)=0.$$ To reduce clutter, I’ll denote the discriminant of this equation by $\Delta = (1-m)^2-4(10-b)$. The solutions to this equation are, of course, $$x = \frac12(m-1)\pm\frac12\sqrt\Delta$$ and substituting this into the parabola’s equation results in $$y=\frac14(m^2+\Delta+39)\pm\frac12m\sqrt\Delta.$$ The square of the length of the corresponding chord is therefore simply $(1+m^2)\Delta$. Setting this equal to $1$ and solving for $b$ yields $$b=10-\frac14(1-m)^2+{1 \over 4(1+m^2)},$$ so you now have the equations of a family of secants with chord length equal to $1$ parameterized by slope, but you really only need $\Delta$ and $b$ to solve the problem.  
There’s a property of parabolas that’s handy for attacking this problem: The area bounded by a parabola and its chord is equal to two-thirds of the area of the bounding paralellogram (see here for details). The problem thus becomes one of finding the value of the slope $m$ that maximizes the area of this paralellogram, which is a lot easier to compute. In fact, since the chord length is fixed at $1$, the problem reduces even further to finding the value of $m$ for which the distance between the chord and the tangent parallel to it is maximized. By symmetry, that’s likely to occur when the tangent is at the vertex, i.e., when $m=0$.  
Either by differentiating or by using the fact that midpoints of parallel chords lie on a line parallel to the parabola’s axis, you can find that the tangent to the parabola has a slope of $m$ when $x=\frac12(m-1)$. Substituting the value of $b$ computed above into the expression for $\Delta$ simplifies it to $\Delta = \frac1{1+m^2}$. The distance between the secant and this tangent is equal to the difference between the $y$-coordinates of the midpoint of the chord and the point of tangency†, divided by the normalizing factor of $\sqrt{1+m^2}$, which simplifies to $$\frac\Delta{4\sqrt{1+m^2}} = \frac1{4(1+m^2)^{3/2}}.$$ Therefore, the area of the parabolic segment is $\frac1{6(1+m^2)^{3/2}}$, which obviously has its maximum at $m=0$, as suspected.

† Alternatively, find the $y$-intercept of the tangent by solving $\Delta=0$ for $b$, and subtract this from the value of $b$ computed above for the secant.
A: The parabola has the same shape as $y=x^2$. The chord length of $1$ parallel to the $x$-axis connects points $\big(\pm\frac 12, \frac14\big)$. By symmetry this chord gives the largest area. 
See Desmos illustration here.
For simplicity, in the illustration, the chord is fixed as the segment between $(0,0)$ and $(1,0)$, and the parabola rotates.




The area bounded by the parabola and the chord is given by 
$$2\int_0^{\frac 12}\frac 14-x^2 \;\;dx=\frac 16$$
