Maximum volume?

[sia.wears.hats]

I have a file 8 inches wide by 20 inches long and am folding it to form a flat U shape. How tall should the height be to maximize the volume it can hold?

I know that the volume=8*x*(20-2x) (right?), but I just can't figure out where to go from here.

Any help is appreciated!

 

[soroban]

Hello, sia.wears.hats!

I have a file 8 inches wide by 20 inches long and am folding it to form a flat U shape. How tall should the height be to maximize the volume it can hold?

I know that the volume = 8*x*(20 - 2x), right? . Right!
But I just can't figure out where to go from here.

I will assume you can't use Calculus.

You have the function: .\(\displaystyle V \;=\;8x(20-2x)\)

That is: .\(\displaystyle V \;=\;-16x^2 + 160x\)

This a parabola that opens downward: \(\displaystyle \frown\)
Its maximum value occurs at its vertex.

Vertex: .\(\displaystyle x \;=\;\dfrac{\text{-}b}{2a} \;=\;\dfrac{\text{-}160}{2(\text{-}16)} \;=\;5\)

For maximum volume, the height should be 5 inches.

 

[sia.wears.hats]

You have the function: .\(\displaystyle V \;=\;8x(20-2x)\)

That is: .\(\displaystyle V \;=\;-16x^2 + 160x\)

This a parabola that opens downward: \(\displaystyle \frown\)
Its maximum value occurs at its vertex.

Vertex: .\(\displaystyle x \;=\;\dfrac{\text{-}b}{2a} \;=\;\dfrac{\text{-}160}{2(\text{-}16)} \;=\;5\)

For maximum volume, the height should be 5 inches.

Oh okay! Thank you Soroban! That makes sense. I hadn't heard of the equation for the vertex yet so that helps on all my other problems too! Thank you so much!

 

[Ishuda]

I have a file 8 inches wide by 20 inches long and am folding it to form a flat U shape. How tall should the height be to maximize the volume it can hold?

I know that the volume=8*x*(20-2x) (right?), but I just can't figure out where to go from here.

Any help is appreciated!

Assuming that open space got closed in some way without taking away any of the file, yes
V(x) = 8*x*(20-2x) = 16 x (10 - x)
where V is the volume depending on the height x. Since the volume is non-negative, x is between 0 and 10 inclusive.

One way to determine the maximum is to use a bit of calculus, i.e. take the derivative of the function and find where it is zero but, since this is posted in the Geometry and Trig section, I would think you wouldn't want to do it that way.

So I would assume that from the geometry part of it, you would know that if a second order equation has the same value at two x values that the extremum is midway between the two x values. In this particular case the extremum is a maximum since the volume at x equal 0 and 10 are zero and the volume is positive between these two values.

 

[stapel]

I hadn't heard of the equation for the vertex yet...

Does your class cover it at all? (Check in the index in the back of your book, probably under "vertex".) If not (or, at least, if not yet), then you should probably use the method they gave you in class, which was probably "completing the square". Especially on the test, if you're supposed to show your work (and especially if you're supposed to use completing the square), the "right" number found by the "wrong" method could be counted off.

If you're not sure, ask your teacher before the next test!