# Using derivatives to find the max

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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Note that this is similar to your other problem. Find an expression for the cross-section (or whatever you're given), try to find another formula you can relate (area for this problem), get things in terms of one variable, find critical points, plug into earlier expressions. – Joe Jun 11 '12 at 2:44
Again, remember: drawing a picture is the best way to get started! – Cameron Buie Jun 11 '12 at 2:58

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.
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. – Brian M. Scott Jun 11 '12 at 3:04