The US Postal service stipulates that any boxes sent through the mail must have a length plus girth totaling no more than 108 inches. Find the dimensions of the box with maximum volume that can be sent through the us mail assuming that the width and height of the box are equal
currently studying applications of extrema if that helps
Your best bet for this would be a cube, so\[108=5s\]Which would give us a cube a little more than 21 inches on each side.
Sorry, jumped to the answer by knowing about how these things work. Let's consider how we will measure this box.
Since the box is square in two dimensions, let the width and height be x. The length will be 108-4x (this comes from the specification of length plus girth).
So the volume is\[V(x)=x^2(108-4x)=108x^2-4x^3\]
Are you with me so far?
yep, do you find the critical point of that then?
Exactly.
sweet, thank you very much
I'm interested to see if my farmer's estimate is correct. So we differentiate\[V'(x)=216x-12x^2=0\rightarrow x=0, x=18\]
I think you probably get the maximum at x=18. I wonder why it's different from my initial estimate?
My calculus answer seems ok, but isn't even a maximum box size. Something is wrong.
I found another solution online that shows the length being 108-x. This gives the equation \[V(x)=x ^{2}(108-x)\] and the derivative of 216x-3x^2. X would then be 72 inches
making y=36
Yeah, but the girth is the distance around. If we have a box that has a square base with side length s, the girth is 4s. That's what needs to be added to the height to get 108
ah ok, yeh you are right then
But my calculus answer is still wrong.
So w = h =18 and l = 108 - 4h = 108 - 4*18 = 108 - 72 = 36 So maximum volume = lwh = 36*18*18 = 11664 cubic inches
Ok, I guess there is something I missed on my initial estimate that it had to be a cube....
Actually, the answer is two cubes stuck end-to-end. Hmmm.....
Nice work, Astro. I hope I was helpful.
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