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Math Help - Applications of Derivatives - optimization problems

  1. #1
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    Applications of Derivatives - optimization problems

    This topic of optimization is tough for me to grasp and having lots of trouble with the problems

    Question
    a rectangle has its base on the x-axis and its upper two vertices on the parabola y = 12-x^2. what is the largest area the rectangle can have, and what are its dimensions ?

    I drew a diagram of a parabola going to the negative and then the rectangle in it
    but now would I set the area of a rectangle A = xy equal to the equation above?
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  2. #2
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    Quote Originally Posted by jddery View Post
    This topic of optimization is tough for me to grasp and having lots of trouble with the problems

    Question
    a rectangle has its base on the x-axis and its upper two vertices on the parabola y = 12-x^2. what is the largest area the rectangle can have, and what are its dimensions ?

    I drew a diagram of a parabola going to the negative and then the rectangle in it
    but now would I set the area of a rectangle A = xy equal to the equation above?
    Ok, so when you draw the diagram you need to try to label the sides of the rectangle. I labeled the vertices on the x-axis as -x and x. So the length of the base is 2x. The rectangles vertices on the parabola are labeled y, so the lenght of the sides is y. So the area, is A(x,y)=2xy. Now substitue the parabola to make this a function of x alone.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by jddery View Post
    This topic of optimization is tough for me to grasp and having lots of trouble with the problems

    Question
    a rectangle has its base on the x-axis and its upper two vertices on the parabola y = 12-x^2. what is the largest area the rectangle can have, and what are its dimensions ?

    I drew a diagram of a parabola going to the negative and then the rectangle in it
    but now would I set the area of a rectangle A = xy equal to the equation above?
    Because of symmetry, the two base vertices are at (\pm x,0) and the two upper vertices are 12-x^2. So the area is \text{base}\times\text{height}=(x-(-x))\cdot(12-x^2)=24x-2x^3

    Take the derivative of this and set it equal to zero to find x that maximizes the area.
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