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diagram
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.

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.

Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thank you Scott. Much appreciated answer and easy to understand.

Algebra

Volunteer

#### Scott A Wilson

##### Expertise

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

##### Experience

I have solved story problems, linear equations, parabolic equations. I have also solved some 3rd order equations and equations with multiple variables.

Publications
Documents at Boeing in assistance on the manufacturiing floor.

Education/Credentials
MS at math OSU in mathematics at OSU, 1986. BS at OSU in mathematical sciences (math, statistics, computer science), 1984.

Awards and Honors
Both my BS and MS degrees were given with honors.

Past/Present Clients
Students in a wide variety of areas since the 80's; over 1,000 of them have been in algebra.