geometrylinear algebrarectanglesvolume
I came across a question:
The surface area of the six faces of a rectangular solid are 4, 4, 8, 8, 18 and 18 square cms. The volume of the solid, in cubic centimetres is __.
I can guess that 4 is $lb$, and 8 is $bh$, and 18 is $lh$. But now if I suppose that $b = 2$, then I get $l = 2$ but it makes $h = 9$ which makes $bh = (2)(9) = 18$ false.
Can anyone show a better working for this.
Thanks a lot.
Hint
From the perimeter, you can extract $l$. From the surface, you can extract $b$. So, now, $V$ is only a function of $h$ and you want to maximize $V$. Then, ...
I am sure that you can take from here.
Here's an explanation that mixes varying amounts of intuition and slight rigor.
Volume of a solid of revolution:
When using the disk method the idea is that we're adding up the volumes of a massive amount of extremely thin disks between $x=a$ and $x=b$ in order to get the volume of the solid. The disks each have radius given by $y(x)$ and thickness given by $\Delta x$. So the volume of each disk is $\pi [y(x)]^2 \Delta x$. Then we push this through the limit process so that we're using integration to add up the volumes of infinitely many infinitely thin disks. Here, $y(x)$ stays $y(x)$ and $\Delta x$ becomes $dx$ in the integral, and we have $\displaystyle V = \pi \int_a^b [y(x)]^2 \, dx$.
The "slant" of the function is completely irrelevant here.
Surface area of a solid of revolution:
To find the surface area, you want to add up the surface areas of the boundaries of a massive amount of extremely tiny approximate disks. (My use of the word "approximate" will be explained shortly, and until then I'll just keep saying disk and I'll also stop specifying that we only want the surface areas of the boundaries.) Each disk has radius given by $y(x)$. But this time, we can't take $\Delta x$ to be the thickness. Why not? Because $\Delta x$ is a good approximation for the thickness of the interior of the disk, but not for the boundary of the disk. And it's the boundary we care about for surface area, because surface area depends on circumference, and the circumference depends on what's happening on the boundary.
How can we get the thickness (or length) of the boundary? We need to approximate the boundary with tiny line segments. Each line segment can be viewed as the hypotenuse of a tiny right triangle whose legs are parallel to the $x$- and $y$-axes. The leg parallel to the $x$-axis has length $\Delta x$ and the leg parallel to the $y$-axis has length $\Delta y$. If we let $\Delta s$ denote the length of our tiny line segment then we get $\Delta s = \sqrt{(\Delta x)^2 + (\Delta y)^2}.$
Here's a picture illustrating this. Imagine this is some function that we very closely zoomed in on:
So, each of our little tiny disks has surface area approximately given by $2\pi y(x) \sqrt{(\Delta x)^2 + (\Delta y)^2}$. Note that our disks actually have slanted boundaries. (The line segments whose lengths are $\Delta s$ are the boundaries of our disks.) This is why I said "approximate disks" earlier: Our tiny disks for surface area are not perfect disks (like we had for volume) because they're actually disks with slanted boundaries. And this whole slanted boundary thing is necessary here because we must accommodate for the slant of the function, because that slant affects the boundaries of our disks (but not their interiors, which is why we didn't care about slant for the volume).
Anyway, we can rewrite the approximate surface area of each slanted-boundary disk as: $$ 2\pi y(x) \sqrt{(\Delta x)^2 + (\Delta y)^2} = 2\pi y(x) \sqrt{(\Delta x)^2 \left[\frac{(\Delta x)^2}{(\Delta x)^2} + \frac{(\Delta y)^2}{(\Delta x)^2}\right]} = 2\pi y(x) \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2} \, \Delta x$$
Then we push this through the limit process so that we're using integration to add up the surface areas of infinitely many infinitely thin disks. Here, $y(x)$ stays $y(x)$, and $\Delta y$ becomes $dy$ and $\Delta x$ becomes $dx$. So we get $\displaystyle S = 2\pi \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$.
Summary:
For volume, we start by adding up the volumes of a massive amount of extremely thin disks. Since we're adding volumes, it's the interiors of the disks that we care about. And $\Delta x$ is a good enough approximation for the thickness of each disk if we're looking only at the interiors of the disks.
For surface area, we start by adding up the surface areas of a massive amount of extremely thin disks. Since we're adding surface areas, it's the boundaries of the disks that we care about. But $\Delta x$ is not a good enough approximation for the thickness of each disk, because it doesn't account for the slant of the function. This is why $\Delta s$ is necessary. And the slant of the function is important for this because the function is the boundary of our solid of revolution, and we need to use the thickness of that boundary to get the circumferences of our "disks" so that we can get a working approximation of the surface areas.
Best Answer
We have $4 \cdot 8 \cdot 18 = lbbhlh$ or $576=l^2b^2h^2$. Taking square roots gives $lbh=24$.