SOLUTION: A rectangle has one corner on the graph of y=36-x^2 another at the origin a 3rd on the positive y-axis, and the fourth on the positive x-axis. Express the area A of the rectangle a
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-> SOLUTION: A rectangle has one corner on the graph of y=36-x^2 another at the origin a 3rd on the positive y-axis, and the fourth on the positive x-axis. Express the area A of the rectangle a
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Question 854501: A rectangle has one corner on the graph of y=36-x^2 another at the origin a 3rd on the positive y-axis, and the fourth on the positive x-axis. Express the area A of the rectangle as a function of x. What is the domain of A? For what value of x is A largest?
ANY HELP WOULD BE SOOOO WELCOME!! Thanks in advance I am super confused! Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! It's a good 1st step to draw the parabola
and also a horizontal line which represents
both the point on the y-axis ( 0, y ) and
also the intersection of this line with the
parabola ( x, y ).
Remember that the horizontal line can be
ANY horizontal line
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So far, I have the points:
( 0, 0 )
( 0, y )
( x, y )
Now I can add the final point ( x1, 0 )
Both x and y are positive
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I can plot this on top of the previous plot
The domain is to
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Are you in a calculus class? That's the only way I know to
find the peak of 3rd degree equation ( slope = 0 )
This is the value of for which is largest
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and now find
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Hope this all makes sense- if you do it with a
non-calculus method, you should get the
same answer
