Max/Min Problems

  • Thread starter Riogho
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  • #1
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Homework Statement


A rectangle is bounded by the X-axis and the semicircle Y = [(sqrt)36-x^2]. What dimensions should the rectangle have so that its area is a maximum.


Homework Equations



Just a note, the 36-x^2 is all under the radical.

The Attempt at a Solution



Ditto.
 

Answers and Replies

  • #2
gabbagabbahey
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Well, the width of the rectangle is clearly [itex]2x[/itex], and its height is [itex]y=\sqrt{36-x^2}[/itex]....so what is its area [itex]A(x)=?[/itex] for a given value of [itex]x[/itex]? How do you find the maximum of such a function?

P.S. I don't think the definition of "ditto" is exactly what you seem to think it is (just a friendly FYI)
 
  • #3
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Wouldn't I just find the area by multiplying both of them together?
 
  • #4
gabbagabbahey
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Sounds good to me; the last time I checked the area of a rectangle was just width times height ;0)
 
  • #5
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So I multiply them together then take the derivative, and then find the critical points of that?
 
  • #6
HallsofIvy
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Yes, now do it!
 
  • #7
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Done.

Many thanks.
 

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