Find the maximum area of a triangle formed in the first quadrant by the xx-axis, yy-axis and a tangent line to the graph of f = ( x + 6 ) − 2

Question

Find the maximum area of a triangle formed in the first quadrant by the xx-axis, yy-axis and a tangent line to the graph of   f  =  (  x  +  6      )          −      2

Gabriella Hensley · Accepted Answer

Solution:  y  =  (  x  +  6      )          −      2                    d      y              d      x        =  −  2  (  x  +  6      )          −      3        =            −      2              (      x      +      6              )        3            At x=a            d      y              d      x        =      2          (      a      +      6              )        3            The equation of tangent at x=a is  y  −      1          (      a      +      6              )        2              =            −      2              (      a      +      6              )        3              (  x  −  a  )    ⇒  y            (      a      +      6              )        2            −      1              (      a      +      6              )        2              =            −      2      (      x      −      a      )              (      a      +      6              )        3                y  (  a  +  6      )    3    −  (  a  +  6  )  =  −  2  (  x  −  a  )    y  (  a  +  6      )    3    +  2  (  x  −  a  )  −  (  a  +  6  )  =  0Set x=0 for y intercept  y  (  a  +  6      )    3    +  2  (  x  −  a  )  −  (  a  +  6  )  =  0    y  =            3      a      +      6              (      a      +      6              )        3            Set y=0 for x-intercept  y  (  a  +  6      )    3    +  2  (  x  −  a  )  −  (  a  +  6  )  =  0    2  (  x  −  a  )  −  (  a  +  6  )  =  0    x  =            3      a      +      6        2  Now,Area of triangle   (  A  )  =      1    2    x  y    =      1    2    (            3      a      +      6        2    )  (            3      a      +      6              (      a      +      6              )        3              )To maximize this area we have to differentiate with respect to a            d      A              d      a        =      d          d      a        (      1    4              (      3      a      +      6              )        2                    (      a      +      6              )        3              )    =      1    4    ⋅            6      (      3      a      +      6      )      (      a      +      6              )        3            −      3      (      a      +      6              )        2            (      3      a      +      6              )        2                    (      (      a      +      6              )        3                    )        2                =            3      (      3      a      +      6      )      (      −      a      +      6      )              4      (      a      +      6              )        4                          d      A              d      a        =  0    3  (  3  a  +  6  )  (  −  a  +  6  )  =  0which gives a=-2,6when a=-2, A=0when a=6, a=1/12The maximum area   =      1    12

Find the maximum area of a triangle formed in the first quadrant by the xx-axis, yy-axis and a tangent line to the graph of f=(x+6)^(−2)

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