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Find the dimensions of the rectangle of maximum area that can be formed from a 40-in. piece of wire
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20 X 2
i already tried that it is incorrect
5 X 8
yep!!!!!
|dw:1436312621888:dw| the perimeter is 40 inches\[l+l+w+w = 40 \implies 2l+2w = 40\] now we can solve for either width or length, so solving for length gives us \[l = \frac{ 40-2w }{ 2 }\]using this information we can find the area of a rectangle since we know area of a rectangle is just \[A = length \times width \]\[A = l \times w\] so plugging our length into the area formula we have \[A = (20-w)w = 20w-w^2\] I simplified the length by dividing by 2 from the initial length formula I made. So now to find the maximum area, we need to take the derivative, I leave the rest to you.
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