The maximum possible area bounded by the parabola $y = x^2 + x + 10$ and a chord of the parabola of length $1$ is?

Question

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 $1$?

Answer 1

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.

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† 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.

Answer 2

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 $y_1 = x^{2}_{1} + x_1 + 10$.

Answer 3

I consider the more general parabola

$y(x) = x^2 + ax + b $ with the points being $(x_i, y_i)_{i=1}^2 $.

I am close to a solution, but am pooping out so am leaving my answer incomplete.

One surprising result I find is that the area between the chord and the parabola is $\dfrac{(x_2-x_1)^3}{6} $.

The length of the chord is

$\begin{array}\\ L^2 &=(x_2-x_1)^2+(y_2-y_1)^2\\ &=(x_2-x_1)^2+((x_2^2 + ax_2 + b)-(x_1^2 + ax_1 + b))^2\\ &=(x_2-x_1)^2+(x_2^2-x_1^2 + a(x_2-x_1))^2\\ &=(x_2-x_1)^2+((x_2-x_1)(x_2+x_1) + a(x_2-x_1))^2\\ &=(x_2-x_1)^2+(x_2-x_1)^2((x_2+x_1) + a)^2\\ &=(x_2-x_1)^2(1+((x_2+x_1) + a)^2)\\ \end{array} $

If $x_2-x_1 = d$, $L^2 =d^2(1+((2x_1+d) + a)^2) $.

The equation of the chord is

$\begin{array}\\ \dfrac{y-y_1}{x-x_1} &=\dfrac{y_2-y_1}{x_2-x_1}\\ &=\dfrac{(x_2^2+ax_2+b)-(x_1^2+ax_1+b)}{x_2-x_1}\\ &=\dfrac{x_2^2-x_1^2+a(x_2-x_1)}{x_2-x_1}\\ &=x_2+x_1+a\\ &=u\\ \end{array} $

where $u = x_2+x_1+a$,

or

$\begin{array}\\ y &=y_1+(x-x_1)u\\ &=ux+x_1^2+ax_1+b-ux_1\\ &=ux+x_1^2+ax_1+b-(x_2+x_1+a)x_1\\ &=ux+b-x_2x_1\\ \end{array} $

The area under the chord is

$\begin{array}\\ A_c &=\int_{x_1}^{x_2} (ux+y_1-ux_1)dx\\ &=(\dfrac{ux^2}{2}+y_1x-ux_1x)|_{x_1}^{x_2}\\ &=\dfrac{u(x_2^2-x_1^2)}{2}+y_1(x_2-x_1)-ux_1(x_2-x_1)\\ &=(x_2-x_1)(\dfrac{u(x_2+x_1)}{2}-ux_1+y_1)\\ &=(x_2-x_1)(\dfrac{u(x_2-x_1)}{2}+y_1)\\ &=(x_2-x_1)(\dfrac{(x_2+x_1+a)(x_2-x_1)}{2}+y_1)\\ &=(x_2-x_1)(\dfrac{x_2^2-x_1^2+a(x_2-x_1)}{2}+x_1^2+ax_1+b)\\ &=(x_2-x_1)(\dfrac{x_2^2+x_1^2+a(x_2+x_1)}{2}+b)\\ \end{array} $

The area under the parabola is

$\begin{array}\\ A_p &=\int_{x_1}^{x_2} (x^2+ax+b)dx\\ &=(\dfrac{x^3}{3}+a\dfrac{x^2}{2}+bx)|_{x_1}^{x_2}\\ &=\dfrac{x_2^3-x_1^3}{3}+a\dfrac{x_2^2-x_1^2}{2}+b(x_2-x_1))\\ &=(x_2-x_1)\left(\dfrac{x_2^2+x_2x_1+x_1^2}{3}+a\dfrac{x_2+x_1}{2}+b\right)\\ \end{array} $

The difference is, noting that the $a$ and $b$ terms cancel out,

$\begin{array}\\ A_{cp} &=A_c-A_p\\ &=(x_2-x_1)\left(\dfrac{x_2^2+x_1^2}{2}-\dfrac{x_2^2+x_2x_1+x_1^2}{3}\right)\\ &=(x_2-x_1)\left(\dfrac{x_2^2+x_1^2-2x_2x_1}{6}\right)\\ &=\dfrac{(x_2-x_1)^3}{6}\\ \end{array} $

I find this very surprising.

This is the kind of result that, if it is really true, probably has a much simpler proof.

Now to incorporate the length restriction.

We have $L^2 =(x_2-x_1)^2(1+((x_2+x_1) + a)^2) $.

So we want to minimize $(x_2-x_1)^3$ subject to $L^2 =(x_2-x_1)^2(1+((x_2+x_1) + a)^2) $.

The following discussion is taken from https://en.wikipedia.org/wiki/Lagrange_multiplier

The Lagrangian is $h(x_1, x_2, \lambda) =(x_2-x_1)^3 -\lambda((x_2-x_1)^2(1+((x_2+x_1) + a)^2)-L^2) $ or, using simplified variables ($x_1 \to x, x_2 \to y, \lambda \to z$), $h(x, y, z) =(y-x)^3 -z((y-x)^2(1+((y+x) + a)^2)-L^2) =f(x, y) -zg(x, y) $.

So we want to solve

$g(x, y) = 0, \nabla_{x, y}f(x, y) =z\nabla_{x, y}g(x, y) $ where $\nabla_{x, y}f(x, y) =\left(\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}\right) $.

I think this is close to the end, but it's late and I'm tired.

So, once again, I am leaving an incomplete answer in the hope that someone else can finish it.

Answer 4

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 $$2\int_0^{\frac 12}\frac 14-x^2 \;\;dx=\frac 16$$