A wire 24 inches long is to be cut into 4 pieces to form a rectangle whose shortest side has a length of x. a)Express the area A of the rectangle as a function of x.
12x -x^2
\[x(12-x)\] i think should work
the answer was a minute late, that should be in picoseconds!
so both would work??
idea is this if x is one side and y the other then the perimeter is \[2x+2y=24\] so \[x+y=12\] and \[y=12-x\] making the area (length times width) \[A(x)=x(12-x)\]
yes they are the same \[A(x)=x(12-x)=12x-x^2\]
ok and it says that i should make a conjecture about the dimensions that yeild a maximum area
make a square (more or less obvious)
i mean it is really obvious, because this is symmetric in x and y so it is biggest when they are equal. need some algebra to prove it?
sure
\[A(x)=12x-x^2\] a parabola that opens down. biggest at the vertex which is at \[-\frac{b}{2a}=-\frac{12}{-2\times-1}=6\] so x should be 6 and therefore y should also be 6 and so it is a square with area = 6^2=36
hope you were not supposed to use calc or something
yea this is the first few pages of calc
what would be the domain?
well you could do the following ridiculous thing 1) \[A(x)=12x-x^2\] 2) \[A'(x)=12-2x\] set \[12-2x=0\] get \[x=6\] but it is alway \[-\frac{b}{2a}\] for a quadratic
the domain if we were to graph that. because it asks us to graph also
well you are making a rectangle right? so negative lengths wouldn't make sense
and clearly x cannot be bigger than 24 because that is all you got
ok......
actually scratch that
x is only one side, so it cannot be bigger than 12
alright
i would say domain is \[0<x<12\] and here is a nice picture http://www.wolframalpha.com/input/?i=y%3D12x-x^2+for+0%3Cx%3C12
there now you must be done with this homework problem because we have beaten it to death
lol yes thanks you were very helpful XD
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