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Higher Math Calculating the Minimum Perimeter of a Triangle

san francisco
Last edited:
Hey there sorry if this is in the wrong threadbut I need some help with this problem I'm stuck on.

Give a point (a,b) with 0 < b < a
determine the minimum perimeter of a triangle with one vertex at (a,b)on the x-axis,
and one on the line y = x.

I hope someone can help/guide me through this problemso I can understand how to do this problem thanks.
 
No triangle exists that minimizes that condition.
Howeverthe infinum of the perimeters of all such triangles = twice the distance from (a,b) to the line y = x.
That's just an application of the triangle inequality after picking the first two points (one is on the linethe other is (a,b)) of any triangle.
 
Hellogfbrd!

Did you make a sketch?


Give a point (a,b) with 0 < b < a
determine the minimum perimeter of a triangle
with one vertex at (a,b)one on the x-axisand one on the line y = x.

Code:
      |
      |                     *
      |                   *
      |               R *
      |               o(y,y)
      |             **  *
      |           * *     *
      |         *  *        * P
      |       *   *           o(a,b)
      |     *    *        *
      |   *     *     *
      | *    Q *  *
  - - * - - - o - - - - - - - - - - - -
      |     (x,0)
      |
Point P has given coordinates P(a,b).
It lies below the 45o-line.

Point Q lies on the x-axis with coordinates Q(x,0).

Point R lies on the line y = x with coordinates R(y,y).


The lengths of the sides of the triangle are:

. . PQ \;=\;\sqrt{(x-a)^2 + b^2}

. . QR \;=\;\sqrt{(x-y)^2 + y^2}

. . PR \;=\;\sqrt{(y-a)^2 + (y-b)^2}


The perimeter of the triangle is:

. . Z \;=\;\sqrt{(x-a)^2 + b^2} + \sqrt{(x-y)^2 + y^2} + \sqrt{(y-a)^2 + (y-b)^2}

And that is the function we must minimize.
 
Apologies - ignore my previous post - it's wrong. I simply read right over the phrase "on the x-axis"and so didn't include that condition.
 

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