Recent content by illjazz

Even though we're not counting the "top" of the box, a side's length will still be x=\sqrt{1200}-2y, will it not? If this is where my error lies, I do not see it. I am, of course. I did have to raise my eyebrow when I calculated V'(0). I got 1200! Obviously can't be right.. that's what my...

How so? I don't think I'm following. I mean I see what you're saying, but I don't see how it affects the volume, which is what I'm after :/ Setting this issue aside, does my process otherwise look correct?

Yes, but as I understand it, that does not change the volume of the box. The volume will be the same, regardless of whether there is a top, no?

The largest area that can be encompassed by a 2d shape? Um.. I'd say infinity. A shape is approximated by taking a 2d object with n "sides" and then increasing the number of sides long enough to reach the desired level of precision, I guess? Ok. So we have the sides, x and y. Then the area A...

Homework Statement (11) a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square. b) Show that of all the rectangles with a given perimeter, the one with greatest area is a square. Homework Equations - Differentiation rules - Superhuman powers...

Homework Statement If 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box. Homework Equations - Differentiation rules - Magic? The Attempt at a Solution First, I drew a square. I drew little squares in...

Got it! Thanks for the tip :) y'=\frac{e^{-x} + (x\frac{d}{dx}e^{-x}+e^{-x}\frac{d}{dx}x)}{e^{-x}+xe^{-x}}) \frac{-e^{-x}-xe^{-x}+e^{-x}}{e^{-x}+xe^{-x}} \frac{e^{-x}(1-1-x)}{e^{-x}+xe^{-x}} \frac{e^{-x}(-x)}{e^{-x}(x+1)} \frac{-x}{x+1} ! :)

Ok I tried the following: y=\ln(e^{-x}+xe^{-x}) y'=\ln e^{-x}+\ln(xe^{-x}) -e^{-x}\ln e+\ln x+\ln e^{-x} -e^{-x}+\ln x-e^{-x}\ln e -e^{-x}+\ln x-e^{-x} -2e^{-x}+\frac{1}{x} -2\frac{1}{e^x}+\frac{1}{x}=-\frac{2}{e^x}+\frac{1}{x} But frankly, I don't know if this is...

Oops.. I assumed that if the derivative of e^x is itself, the x stood as a placeholder for, well, pretty much anything, including -x. So um.. what does e^(-x) come out to be? Ah wait.. just saw it. If y = e^(-x), then y' = e^(-x)ln(e) Hmm.. is that correct? This is the rule for...

Ok I looked at this a second time and got a bit further.. I'm still missing the solution though. \frac{e^{-x}+xe^{-x}+e^{-x}}{e^{-x}+xe^{-x}}=\frac{2e^{-x}+xe^{-x}}{e^{-x}+xe^{-x}} \frac{\frac{1}{e^{x}}+\frac{x}{e^{x}}+\frac{1}{e^{x}}}{\frac{1}{e^{x}}+\frac{x}{e^{x}}}...

Ah! Brilliant! Thank you! So I did get it :). Much appreciated.

Homework Statement Differentiate y=\ln(e^{-x} + xe^{-x}) Homework Equations - Logarithmic differentiation? The Attempt at a Solution Here goes: y=\ln(e^{-x} + xe^{-x}) y'=\frac{e^{-x} + (x\frac{d}{dx}e^{-x}+e^{-x}\frac{d}{dx}x)}{e^{-x}+xe^{-x}})...

Thanks! I actually did not know that.. logs aren't my strength :/ That helps with the numerator, but not with the denumerator. Continuing from the last step: \frac{1+\ln 2 + \ln u-\ln u}{u(1+2\ln 2+2\ln u + (\ln 2 + \ln u)^2)} \frac{1+\ln 2}{u(1+2\ln 2+2\ln u + (\ln 2 + \ln u)^2)}

Homework Statement Differentiate the function f(u)=\frac{\ln u}{1+\ln(2u)} Homework Equations - Quotient rule? - Logarithmic differentiation? The Attempt at a Solution I just learned about logarithmic differentiation so I think the idea here was to use logarithmic...

Absolutely! I guess what threw me of is what I'd initially written.. ending up at equation = 0. Something not existing, aka being undefined, does not mean it is equal to 0, as we know.. I'll have to get back to you later on that "proof" because I'm working on a bunch of other problems as...