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Algebra/quadratic equation


I have a piece of metal 10m long and 0.3m wide. I need to bend it into an open U shaped channel (gutter) by bending the 0.3m width twice. I need to use a quadratic equation to to determine where to make the bends to get the maximum volume.
Thankyou for your help. Peter

If the bottom has width x, the height is (0.3-x)/2.
This makes the cross sectional area into x(0.3-x)/2 = 0.15x - 0.5x^2.
Let this be f(x).

Maximize by finding f'(x), setting it to 0, and solving for x.
Note that here f'(x) = 0.15 - x, so the solution is x = 0.15.
This makes the height into (0.3-0.15)/2 = 0.075.

The cross sectional area is then 0.15 x 0.075 = 0.01125.


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Scott A Wilson


Any algebraic question you've got. That includes question that are linear, quadratic, exponential, etc.


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