areacalculuscalculus-of-variationsconic sectionseducation
This is not a homework question.
I am a new teacher (just graduated) and a student asked me this question.
The points $A(3,9)$ and $B(-2,4)$ lie on the parabola $y=x^2.$ The line $y=x+6$ joins $A$ and $B$. The point $P(p,p^2)$ is a variable point on the parabola below the line.
Show that the greatest possible area of the triangle $APB$ is three-quarters of the area of the parabolic segment $APB$, given that the area of parabolic segment is integral from $-2$ to $3$ of $x^2 dx$.
$\int_a^b f(x)$ gives the area "under the curve" of $f(x)$ between $a$ and $b$: the area from the $x$-axis to the curve, across that interval.
In the cases given, you have two curves that you're dealing with instead; one (which I'll call $f(x)$) is higher and the other ($g(x)$) is lower. Finding the area between these curves means finding the area that is under $f$ but not under $g$. This, it is simple to see, we can do by subtraction: the vertical space between $f(x) = \sqrt{x}$ and $g(x) = x^2$ at, say, $1/4$ is equal to $$f\left(\frac{1}{4}\right)-g\left(\frac{1}{4}\right) = \frac{1}{2} - \frac{1}{16} = \frac{7}{16}$$ because while there's $1/2$ below $f$ there, $1/16$ is also below $g$ and thus shouldn't get counted.
Here's this pair of functions with their integrals displayed, overlapping. The vertical space mentioned above is also shown as a vertical line.
The area under $f$ contains both the pale area and the darker area; the area under $g$ contains only the darker area. But we need only the pale area; we can find both (by taking the integral of $f$), and then remove the dark area (by taking the integral of $g$ and subtracting that from the integral of $f$).
There seems to be a typo in the problem : The $V = 2\sqrt{gd}$ and then the rest follows as given below in the solution. I think I got what the book answer is.
Solution as follows:
Best Answer
We can show this by computation and comparison of areas.
The line through the points $(-2,4)$ and $(p,p^2)$ is $$y_1=(p-2)x+2p,$$ and the line through $(p,p^2)$ and $(3,9)$ is $$y_2=(p+3)x-3p.$$ (used the basic method here of course, and we may assume that $-2<p<3$.)
Now we compute the area of the triangle as the sum of two integrals, each between the line $y=x+6$ and our found lines in terms of $p$ over the appropriate bounds. $$\int_{-2}^p (x+6-((p-2)x+2p))\, dx+\int_{p}^3 (x+6-((p+3)x-3p))\, dx=-\frac{5p^2}{2}+\frac{5p}{2}+15.$$ (I will write the laborious steps if you want, but I expect you have that down pat.)
So we have an area function to maximize in $p$, namely, $$A(p)=-\frac{5p^2}{2}+\frac{5p}{2}+15.$$ We can take the cheap way out here, and recognize that the vertex occurs at $$p=\frac{-b}{2a}=\frac{1}{2},$$ and so the maximum value occurs at $$A\left( \frac{1}{2} \right)=\frac{125}{8}.$$
Now the area of the parabolic segment $APB$ is the area between the line $y=x+6$ and the parabola $y=x^2$ on the bound $-2\leq x \leq 3$. That is, $$\int_{-2}^3 (x + 6 - x^2) \, dx=\frac{125}{6}.$$
We conclude this effort by noting that $$\frac{3}{4}\cdot\frac{125}{6}=\frac{125}{8}.$$
I am more than happy to put the basic algebra steps and polynomial integration steps in (I am staring at them), but your post suggests you already possess those skills. Let me know, and eyes out for typos by all.