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The cross-sectional area of a beam cut from a log with radius 1 foot is given by the function A(x) = 4x•sqrt{1 - x^2}, where x represents the length , in feet, of half the base of the beam.
A. Which value of x maximizes the cross-sectional area?
B. What should be the length of the base of the beam to maximize the cross-sectional area?
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I've used the graph plotter to view what the graph of A against x looks like. I don't think you'll get the exact maximum without calculus but you can zoom in on the point using the plotter and get a value that way.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I've used the graph plotter to view what the graph of A against x looks like. I don't think you'll get the exact maximum without calculus but you can zoom in on the point using the plotter and get a value that way.
Bob
Show me the Calculus way. I know a little bit of Calculus.
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The calculus way is based on the knowledge of the derivatives of functions.
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That will be zero when
Then solve that quadratic.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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That will be zero when
Then solve that quadratic.
Bob
I see. You used the product rule. I am familiar with this rule.
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The calculus way is based on the knowledge of the derivatives of functions.
I am familiar with basic differentiation. I just didn't know that the product rule can be used to solve this application.
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