Find the dimensions of the rectangle of maximum area that can be formed from a 40-in. piece of wire

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Find the dimensions of the rectangle of maximum area that can be formed from a 40-in. piece of wire

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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20 X 2
i already tried that it is incorrect
5 X 8

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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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