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I had some trouble figuring out how to solve this problem. Can anyone help?

A trough with a rectangular cross section is to be made from a long sheet of metal 24 meters wide by turning up strips along each side. Find the amount that must be turned up to give the greatest cross section.

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    Again, remember: drawing a picture is the best way to get started!2012-06-11

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The cross sectional area is $wh$ where $w$ is the width and $h$ is the height. The constraint is $2h+w = 24$. A simple way to do this is to note that $w = 24-2h$. Substitute this into the equation for the area, giving $24h-2h^2$.

To find the maximum area, differentiate the area expression and look for zeros, giving $24-4h=0$. This gives $h=6$, substituting this into the expression for $w$ gives $w=12$.

Alternatively, you could notice that you can write $24h-2h^2 = 72-2(h-6)^2$, from which is is obvious that setting $h=6$ maximizes the expression. I used 'completing the square' to get this alternative expression.

To answer the question, $6m$ ($=h$) must be turned up on either side.

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    Or, if pictorially oriented, the OP could notice that the graph of $A=2h(12-h)$ is a parabola; the parabola opens down, so the vertex is a maximum. The $h$-intercepts are at $0$ and $12$, so the axis of the parabola must be $h=6$, midway between them, and of course the vertex is on the axis.2012-06-11